$=x^{3}-3 x^{2}+x+1$ and its roots are $(a-b)$, a and $(a+b)$. Comparing the given polynomial with $\mathrm{Ax}^{3}+\mathrm{Bx}^{2}+\mathrm{Cx}+\mathrm{D}$, we have: $A=1, B=-3, C=1$ and $D=1$...
Find a quadratic polynomial whose zeroes are 2 and .
Since, the two roots of the polynomial are 2 and $-5$. Let $\boldsymbol{\alpha}=2$ and $\boldsymbol{\beta}=-5$ the sum of the zeroes, $\boldsymbol{\alpha}+\boldsymbol{\beta}=2+(-5)=-3$ Product of...
If one zero of the polynomial is the reciprocal of the other, find the value of a.
$(a+9) x^{2}-13 x+6 a=0$ Here, $A=\left(a^{2}+9\right), B=13$ and $C=6$ a Let $\boldsymbol{\alpha}$ and $\frac{\mathbf{1}}{\boldsymbol{\alpha}}$ be the two zeroes. Then, product of the zeroes...
Find the zeroes of the polynomial .
$p(x)=x^{2}+2 x-195$ Let $p(x)=0$ $\Rightarrow x^{2}+(15-13) x-195=0$ $\Rightarrow x^{2}+15 x-13 x-195=0$ $\Rightarrow \mathrm{x}(\mathrm{x}+15)-13(\mathrm{x}+15)=0$ $\Rightarrow(x+15)(x-13)=0$...
It is given that the difference between the zeroes of is 4 and . Then, (a) (b) (c) (d)
The correct option is option (c) $\frac{5}{2}$ Let the zeroes of the polynomial be $\boldsymbol{\alpha}$ and $\boldsymbol{\alpha}+4$ $p(x)=4 x^{2}-8 k x+9$ Comparing the given polynomial with $a...
If are the zeros of is equal then ? (a) (b) (c) (d)
The correct option is option (c) $\frac{2}{3}$ $p(x)=x^{2}-2 x+3 k$ Comparing the given polynomial with $a x^{2}+b x+c$, we get: $a=1, b=-2$ and $c=3 k$ It is given that $\boldsymbol{\alpha}$ and...
If be the zeroes of the polynomial , then the values of (a) (b) 1 (c) (d) 3
The correct option is option (a) $-1$ $p(x)=x^{3}-6 x^{2}-x+3$ Comparing the given polynomial with $\mathrm{x}^{3}-(\boldsymbol{\alpha}+\boldsymbol{\beta}+\gamma) \mathrm{x}^{2}+(\boldsymbol{\alpha}...
The zeroes of the polynomial are (a) (b) (c) (d) 3,1
(c) $3,-1$ Here, $p(x)=x^{2}-2 x-3$ Let $x^{2}-2 x-3=0$ $\Rightarrow x^{2}-(3-1) x-3=0$ $\Rightarrow x^{2}-3 x+x-3=0$ $\Rightarrow \mathrm{x}(\mathrm{x}-3)+1(\mathrm{x}-3)=0$...
Which of the following is a true statement? (a) is a linear polynomial. (b) is a binomial (c) is a monomial (d) is a monomial
The correct option is option (d) $5 \mathrm{x}^{2}$ is a monomial. $5 \mathrm{x}^{2}$ consists of one term only. So, it is a monomial.
On dividing a polynomial by a non-zero polynomial be the quotient and be the remainder, then , where (a) always (b) always (c) either or (d)
The correct option is (c) either $\mathrm{r}(\mathrm{x})=0$ or $\operatorname{deg} \mathrm{r}(\mathrm{x})<\operatorname{deg} \mathrm{g}(\mathrm{x})$ By division algorithm on polynomials, either...
If be the zeroes of the polynomial such that , then ? (a) 3 (b) (c) -2 (d) 2
The correct option is option (d) 2 $\alpha$ and $\beta$ are the zeroes of $2 \mathrm{x}^{2}+5 \mathrm{x}+\mathrm{k}$, we have: $\alpha+\beta=\frac{-5}{2}$ and $\alpha \beta=\frac{k}{2}$ it is given...
If one of the zeroes of the cubic polynomial is , then the product of the other two zeroes is (a) (b) (c) (d)
The correct option is option (c) $1-\mathrm{a}+\mathrm{b}$ $-1$ is a zero of $x^{3}+a x^{2}+b x+c$, we have $(-1)^{3}+a \times(-1)^{2}+b \times(-1)+c=0$ $\Rightarrow a-b+c+1=0$ $\Rightarrow...
If one of the zeroes of the cubic polynomial is 0 , then the product of the other two zeroes is (a) (b) (c) 0 (d)
The correct option is option (b) $\frac{c}{a}$ $\alpha, \beta$ and 0 be the zeroes of $a x^{3}+b x^{2}+c x+d$. Then, sum of the products of zeroes taking two at a time is given by $(\alpha...
If two of the zeroes of the cubic polynomial are 0 , then the third zero is (a) (b) (c) (d)
The correct option is option (a) $\frac{-b}{a}$ $\alpha, 0$ and 0 be the zeroes of $a x^{3}+b x^{2}+c x+d=0$ Then the sum of zeroes $=\frac{-b}{a}$ $\Rightarrow \alpha+0+0=\frac{-b}{a}$ $\Rightarrow...
If be the zeroes of the polynomial such that and , then (a) (b) (c) (d) none of these
The correct option is option (c) $x^{3}-3 x^{2}-10 x+24$ $\alpha, \beta$ and $\gamma$ are the zeroes of polynomial $p(x)$. $(\alpha+\beta+\gamma)=3,(\alpha \beta+\beta \gamma+\gamma \alpha)=-10$ and...
If be the zeroes of the polynomial , then ? (a) (b) 3 (c) (d)
The correct option is option (a) $-3$ $\alpha, \beta$ and $\gamma$ are the zeroes of $2 \mathrm{x}^{3}+\mathrm{x}^{2}-13 \mathrm{x}+6$, we have: $\alpha \beta \gamma=\frac{-(\text { constant term...
If be the zeroes of the polynomial , then (a) (b) 1 (c) (d) 30
The correct option is option (a) $-1$ It is given that $\alpha, \beta$ and $\gamma$ are the zeroes of $x^{3}-6 x^{2}-x+30$ $\therefore(\alpha \beta+\beta \gamma+\gamma \alpha)=\frac{\text {...
If are the zeroes of the polynomial , then ? (a) 3 (b) (c) 12 (d)
The correct option is option (b) $-3$ $\alpha$ and $\beta$ be the zeroes of $x^{2}+6 x+2$, we have: $\alpha+\beta=-6$ and $\alpha \beta=2$...
If the sum of the zeroes of the quadratic polynomial is equal to the product of its zeroes, then ? (a) (b) (c) (d)
The correct option is option (d) $\frac{-2}{3}$ $\alpha$ and $\beta$ be the zeroes of $\mathrm{kx}^{2}+2 \mathrm{x}+3 \mathrm{k}$. Then $\alpha+\beta=\frac{-2}{k}$ and $\alpha \beta=3$ $\Rightarrow...
If one zero of be the reciprocal of the other, then ? (a) 3 (b) (c) (d)
The correct option is option (a) $\mathrm{k}=3$ $\alpha$ and $\frac{1}{\alpha}$ be the zeroes of $3 x^{2}-8 x+k$. Then the product of zeroes $=\frac{k}{3}$ $\Rightarrow \alpha \times...
If and 3 are the zeroes of the quadratic polynomial , then (a) (b) (c) (d)
The correct option is option (c) $a=-2, b=-6$ $-2$ and 3 are the zeroes of $x^{2}+(a+1) x+b$. $(-2)^{2}+(a+1) \times(-2)+b=0 \Rightarrow 4-2 a-2+b=0$ $\Rightarrow b-2 a=-2$ ….(1) $3^{2}+(a+1) \times...
If one zero of the quadratic polynomial is , then the value of is (a) (b) (c) (d)
The correct option is option (b) $\frac{5}{4}$ Since $-4$ is a zero of $(k-1) x^{2}+k x+1$ $(k-1) \times(-4)^{2}+k \times(-4)+1=0$ $\Rightarrow 16 \mathrm{k}-16-4 \mathrm{k}+1=0$ $\Rightarrow 12...
If one zero of the quadratic polynomial is 2 , then the value of is (a) (b) (c) (d)
The correct option is option (d) $\frac{-6}{5}$ Since 2 is a zero of $k x^{2}+3 x+k$, we have: $\mathrm{k} \times(2)^{2}+3(2)+\mathrm{k}=0$ $\Rightarrow 4 \mathrm{k}+\mathrm{k}+6=0$ $\Rightarrow 5...
If and are the zeroes of , then the value of is (a) (b) (c) (d)
The correct option is option (c) $\frac{-9}{2}$ $\alpha$ and $\beta$ be the zeroes of $2 \mathrm{x}^{2}+5 \mathrm{x}-9$. If $\alpha+\beta$ are the zeroes, then $\mathrm{x}^{2}-(\alpha+\beta)...
The zeroes of the quadratic polynomial are (a) both positive (b) both negative (c) one positive and one negative (d) both equal
The correct option is option (b) both negative $\alpha$ and $\beta$ be the zeroes of $\mathrm{x}^{2}+88 \mathrm{x}+125$ Then $\alpha+\beta=-88$ and $\alpha \times \beta=125$ This can only happen...
A quadratic polynomial whose zeroes are and , is (a) (b) (c) (d)
The correct option is option (d) $x^{2}-\frac{1}{10} x-\frac{3}{10}$ Since, the zeroes are $\frac{3}{5}$ and $\frac{-1}{2}$ $\alpha=\frac{3}{5}$ and $\beta=\frac{-1}{2}$ sum of the zeroes,...
A quadratic polynomial whose zeroes are 5 and -3, is (a) (b) (c) (d) none of these
(c) $x^{2}-2 x-15$ Since, the zeroes are 5 and $-3$. $\alpha=5$ and $\beta=-3$ Therefore, sum of the zeroes, $\alpha+\beta=5+(-3)=2$ product of the zeroes, $\alpha \beta=5 \times(-3)=-15$ The...
The sum and product of the zeroes of a quadratic polynomial are 3 and respectively. The quadratic polynomial is (a) (b) (c) (d)
The correct option is option (c) $x^{2}-3 x-10$ Sum of zeroes, $\alpha+\beta=3$ Also, product of zeroes, $\alpha \beta=-10$ $\therefore$ Required polynomial $=\mathrm{x}^{2}-(\alpha+\beta)+\alpha...
The zeros of the polynomial are (a) (a) (c) (d) none of these
The correct option is option (a) $\frac{2}{3}, \frac{-1}{7}$ $f(x)=7 x^{2}-\frac{11}{3} x-\frac{2}{3}=0$ $\Rightarrow 21 \mathrm{x}^{2}-11 \mathrm{x}-2=0$ $\Rightarrow 21 \mathrm{x}^{2}-14...
The zeros of the polynomial are (a) (b) (c) (d) none of these
The correct option is option (b) $\frac{-3}{2}, \frac{4}{3}$ $f(x)=x^{2}+\frac{1}{6} x-2=0$ $\Rightarrow 6 \mathrm{x}^{2}+\mathrm{x}-12=0$ $\Rightarrow 6 x^{2}+9 x-8 x-12=0$ $\Rightarrow 3 x(2...
The zeroes of the polynomial are (a) (b) (c) (d) none of these
The correct option is option (c) $\frac{3}{\sqrt{2}}, \frac{\sqrt{2}}{4}$ $f(x)=4 x^{2}+5 \sqrt{2} x-3=0$ $\Rightarrow 4 x^{2}+6 \sqrt{2} x-\sqrt{2} x-3=0$ $\Rightarrow 2 \sqrt{2}...
The zeroes of the polynomial are (a) (b) (c) (d)
The correct option is option (b) $3 \sqrt{2},-2 \sqrt{2}$ $f(x)=x^{2}-\sqrt{2} x-12=0$ $\Rightarrow x^{2}-3 \sqrt{2} x+2 \sqrt{2} x-12=0$ $\Rightarrow x(x-3 \sqrt{2})+2 \sqrt{2}(x-3 \sqrt{2})=0$...
The Zeroes of the polynomial are (a) (b) (c) (d) 3,1
The correct option is option (c) $3,-1$ Let $f(x)=x^{2}-2 x-3=0$ $=x^{2}-3 x+x-3=0$ $=x(x-3)+1(x-3)=0$ $=(x-3)(x+1)=0$ $\Rightarrow \mathrm{x}=3$ or $\mathrm{x}=-1$
Which of the following is not a polynomial? (a) (b) (c) (d)
The correct option is option (d) $x+\frac{3}{x}$ is not a polynomial. It is because in the second term, the degree of $x$ is $-1$ and an expression with a negative degree is not a polynomial.
Which of the following is a polynomial? (a) (b) (c) (d) None of these
The correct option is option (d) none of these A polynomial in $x$ of degree $n$ is an expression of the form $p(x)=a_{0}+a_{1} x+a_{2} x^{2}+\ldots \ldots+$ $a_{n} x^{n}$, where $a_{n} \neq 0$.
If the zeroes of the polynomial are , a and , find the values of a and b.
using the relationship between the zeroes of he quadratic polynomial. Sum of zeroes $=\frac{-\left(\text { coef ficient of } x^{2}\right)}{\text { coefficient of } x^{3}}$ $\therefore...
If are the zeroes of the polynomial , then .
using the relationship between the zeroes of the quadratic polynomial. We have Sum of zeroes $=\frac{-(\text { coef ficient of } x)}{\text { coefficient of } x^{2}}$ and Product of zeroes...
If are the zeroes of the polynomial , then
using the relationship between the zeroes of he quadratic polynomial. Sum of zeroes $=\frac{-\text { (coef ficient of } x)}{\text { coefficient of } x^{2}}$ and Product of zeroes $=\frac{\text {...
If and are the zeros of the polynomial find the value of
using the relationship between the zeroes of the quadratic polynomial. Sum of zeroes $=\frac{-(\text { coef ficient of } x)}{\text { coefficient of } x^{2}}$ and Product of zeroes $=\frac{\text {...
If are the zeroes of the polynomial such that , find the value of
using the relationship between the zeroes of the quadratic polynomial. Sum of zeroes $=\frac{-(\text { coef ficient of } x)}{\text { coefficient of } x^{2}}$ and Product of zeroes $=\frac{\text {...
Find the value of k for which the system of linear equations has an infinite number of solutions:
Solution: The system of equations: $\begin{array}{l} \mathrm{kx}+3 \mathrm{y}=(2 \mathrm{k}+1) \\ \Rightarrow \mathrm{kx}+3 \mathrm{y}-(2 \mathrm{k}+1)=0\dots \dots(i) \end{array}$ And,...
Find the value of k for which the system of linear equations has an infinite number of solutions:
Solution: The system of equations: $2 \mathrm{x}+(\mathrm{k}-2) \mathrm{y}=\mathrm{k}$ $\Rightarrow 2 \mathrm{x}+(\mathrm{k}-2) \mathrm{y}-\mathrm{k}=0\dots(i)$ And, $6 \mathrm{x}+(2 \mathrm{k}-1)...
Find the value of k for which the system of linear equations has an infinite number of solutions:
Solution: The given system of equations: $\begin{array}{l} 2 \mathrm{x}+3 \mathrm{y}=7 \\ \Rightarrow 2 \mathrm{x}+3 \mathrm{y}-7=0\dots \dots(i) \end{array}$ And, $(\mathrm{k}-1)...
For what value of k, the system of equations
Have (i) a unique solution, (ii) no solution? Also, show that there is no value of for which the given system of equation has infinitely namely solutions
Solution: The given system of equations: $\begin{array}{l} x+2 y=3 \\ \Rightarrow x+2 y-3=0\dots \dots(i) \end{array}$ And, $5 x+k y+7=0\dots \dots(ii)$ The equations are of the following form:...
For what value of k, the system of equations
has (i) a unique solution, (ii) no solution?
Solution: The given system of equations are: $\begin{array}{l} x+2 y=5 \\ \Rightarrow x+2 y-5=0\dots \dots(i) \\ 3 x+k y+15=0\dots \dots(ii) \end{array}$ The equations are of the forms: $a_{1}...
For what value of k, the system of equations
has (i) a unique solution, (ii) no solution?
Solution: Given system of equations: $\begin{array}{l} \mathrm{kx}+2 \mathrm{y}=5 \\ \Rightarrow \mathrm{kx}+2 \mathrm{y}-5=0\dots \dots(i) \\ 3 \mathrm{x}-4 \mathrm{y}=10 \\ \Rightarrow 3...
Show that the system of equations
has no solution.
Solution: We can write the system of equations as $6 x+5 y-11=0\dots \dots(i)$ $\Rightarrow 9 \mathrm{x}+\frac{15}{2} \mathrm{y}-21=0\dots \dots(ii)$ The given system is of the form $a_{1} x+b_{1}...
Show that the system equations
has an infinite number of solutions
Solution: The system of equations: $\begin{array}{l} 2 x-3 y=5 \\ \Rightarrow 2 x-3 y-5=0\dots \dots(i) \\ 6 x-9 y=15 \\ \Rightarrow 6 x-9 y-15=0\dots \dots(ii) \end{array}$ The equations are of the...
Find the value of k for which the system of equations has a unique solution:
Solution: The system of given equations: $\mathrm{kx}+3 \mathrm{y}=(\mathrm{k}-3)$ $\Rightarrow \mathrm{kx}+3 \mathrm{y}-(\mathrm{k}-3)-0 \quad \ldots .(\mathrm{i})$ And, $12...
Find the value of k for which the system of equations has a unique solution:
Solution: The system of given equations are $\begin{array}{l} 4 x-5 y=k \\ \Rightarrow 4 x-5 y-k=0 \quad \text {....(i) } \end{array}$ And, $2 x-3 y=12$ $\Rightarrow 2 x-3 y-12=0\dots \dots(ii)$ The...
Find the value of k for which the system of equations has a unique solution:
Solution: The system of equations given are $\begin{array}{l} 4 x+k y+8=0 \\ x+y+1=0 \end{array}$ This given system is of the form: $\begin{array}{l} a_{1} x+b_{1} y+c_{1}=0 \\ a_{2} x+b_{2}...
Find the value of k for which the system of equations has a unique solution:
Solution: The system of equations given are $\begin{array}{l} 5 x-7 y-5=0 \\ 2 x+k y-1=0 \end{array}$ This given system is of the form: $a_{1} x+b_{1} y+c_{1}=0$ $a_{2} x+b_{2} y+c_{2}=0$ where,...
Find the value of k for which the system of equations has a unique solution:
Solution: The system of equations given are $\begin{array}{l} x-k y-2=0 \\ 3 x+2 y+5=0 \end{array}$ This system of equations given is of the form: $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2}...
Find the value of k for which the system of equations has a unique solution:
Solution: The system of equations given are $\begin{array}{l} 2 x+3 y-5=0 \\ k x-6 y-8=0 \end{array}$ The given system is of the form: $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y+c_{2}=0$ where,...
Show that the following system of equations has a unique solution: , Also, find the solution of the given system of equations.
Solution: The system of equations given is: $\begin{array}{l} \frac{x}{3}+\frac{y}{2}=3 \\ \Rightarrow \frac{2 x+3 y}{6}=3 \\ 2 x+3 y=18 \\ \Rightarrow 2 x+3 y-18=0\dots \dots(i) \end{array}$ and...
Show that the following system of equations has a unique solution:
, Also, find the solution of the given system of equations.
Solution: The given system of eq. is: $\begin{array}{l} 2 x-3 y-17=0\dots \dots(i) \\ 4 x+y-13=\dots \dots(ii) \end{array}$ The given eq. are of the form $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2}...
Show that the following system of equations has a unique solution:
,
. Also, find the solution of the given system of equations.
Solution: The given system of equations is $3 x+5 y=12$ $5 x+3 y=4$ These equations are of the forms: $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y+$ where, $a_{1}=3, b_{1}=5$ $a_{1} x+b_{1}...
Solve the system of equations by using the method of cross multiplication:
,
, where and
Solution: On substituting $\frac{1}{x}=u$ and $\frac{1}{y}=v$ in the equations given, we obtain $\mathrm{au}-\mathrm{bv}+0=0$ $a b^{2} u+a^{2} b v-\left(a^{2}+b^{2}\right)=0$ Here, $a_{1}=a,...
Solve the system of equations by using the method of cross multiplication:
Solution: We can write the given equation as: $\begin{array}{l} 2 a x+3 b y-(a+2 b)=0\dots \dots(i) \\ 3 a x+2 b y-(2 a+b)=0\dots \dots(i) \end{array}$ Here, $a_{1}=2 \mathrm{a}, \mathrm{b}_{1}=3...
Solve the system of equations by using the method of cross multiplication:
,
Solution: We can write the given equations as: $\begin{array}{l} \frac{a x}{b}-\frac{b y}{a}-(a+b)=0\dots \dots(i) \\ a x-b y-2 a b=0\dots \dots(ii) \end{array}$ Here, $a_{1}=\frac{a}{b},...
Solve the system of equations by using the method of cross multiplication:
,
Solution: Taking $\frac{1}{x+y}=u$ and $\frac{1}{x-y}=v$, the given equations become: $5 u-2 v+1=0\dots \dots(i)$ $15 u+7 v-10=0 \quad \ldots \ldots($ ii $)$ Here, $\mathrm{a}_{1}=5,...
Solve the system of equations by using the method of cross multiplication:
,
Solution: Taking $\frac{1}{x}=u$ and $\frac{1}{y}=v$, the given equations become: $\begin{array}{l} \mathrm{u}+\mathrm{v}=7 \\ 2 \mathrm{u}+3 \mathrm{v}=17 \end{array}$ We can write the given...
Solve the system of equations by using the method of cross multiplication:
,
Solution: The given equations may be written as: $\begin{array}{l} \frac{x}{6}+\frac{y}{15}-4=0\dots \dots(i) \\ \frac{x}{3}-\frac{y}{12}-\frac{19}{4}=0\dots \dots(ii) \end{array}$ Here...
Solve the system of equations by using the method of cross multiplication:
Solution: The given equations may be written as: $\begin{array}{l} 7 x-2 y-3=0\dots \dots(i) \\ 11 x-\frac{3}{2} y-8=0\dots \dots(ii) \end{array}$ Here $a_{1}=7, b_{1}=-2, c_{1}=-3, a_{2}=11,...
By using the method of completing the square, show that the equation has no real roots.
$2 x^{2}+x+4=0$ $\Rightarrow 4 x^{2}+2 x+8=0 \quad$ (Multiplying both sides by 2) $\Rightarrow 4 x^{2}+2 x=-8$ $\Rightarrow(2 x)^{2}+2 \times 2 x \times...
Solve the system of equations by using the method of cross multiplication:
Solution: The given equations are: $\begin{array}{ll} 3 \mathrm{x}-2 \mathrm{y}+3=0 & \ldots \ldots (i)\\ 4 \mathrm{x}+3 \mathrm{y}-47=0 & \ldots \ldots (ii) \end{array}$ Here $a_{1}=3,...
Solve the system of equations by using the method of cross multiplication:
Solution: The given equations are: $\begin{array}{l} x+2 y+1=0\dots \dots (i) \\ 2 x-3 y-12=0\dots \dots(ii) \end{array}$ Here $a_{1}=1, b_{1}=2, c_{1}=1, a_{2}=2, b_{2}=-3$ and $c_{2}=-12$ On cross...
Find the roots of the given equation:
$\sqrt{3} x^{2}+10 x+7 \sqrt{3}=0$ $\Rightarrow 3 x^{2}+10 \sqrt{3} x+21=0 \quad$ (Multiplying both sides by $\left.\sqrt{3}\right)$ $\Rightarrow 3 x^{2}+10 \sqrt{3} x=-21$ $\Rightarrow(\sqrt{3}...
Solve for x and y :
,
Solution: The given equations are $\begin{array}{l} \frac{x}{a}+\frac{y}{b}=\mathrm{a}+\mathrm{b}\dots \dots(i) \\ \frac{x}{a^{2}}+\frac{y}{b^{2}}=2\dots \dots(ii) \end{array}$ Multiplying...
Find the roots of the given equation:
$\sqrt{2} x^{2}-3 x-2 \sqrt{2}=0$ $\Rightarrow 2 x^{2}-3 \sqrt{2} x-4=0 \quad$ (Multiplying both sides by $\sqrt{2}$ ) $\Rightarrow 2 x^{2}-3 \sqrt{2} x=4$ $\Rightarrow(\sqrt{2} x)^{2}-2 \times...
Solve for x and y :
,
Solution: The given equations are $\begin{array}{l} a^{2} x+b^{2} y=c^{2}\dots \dots (i) \\ b^{2} x+a^{2} y=d^{2} \dots \dots(ii) \end{array}$ Multiplying equation(i) by $\mathrm{a}^{2}$ and...
Solve for x and y :
,
Solution: The given equations are $\begin{array}{l} \mathrm{x}+\mathrm{y}=\mathrm{a}+\mathrm{b} \quad\{\ldots \ldots(i) \\ \mathrm{ax}-\mathrm{by}=\mathrm{a}^{2}-\mathrm{b}^{2}+\ldots \ldots(ii) \\...
Find the roots of the given equation:
$\begin{array}{l} x^{2}-(\sqrt{2}+1) x+\sqrt{2}=0 \\ \Rightarrow x^{2}-(\sqrt{2}+1) x=-\sqrt{2} \\ \Rightarrow x^{2}-2 \times x...
Solve for x and y :
,
Solution: The given eq. are: $\frac{b x}{a}+\frac{a y}{b}=\mathrm{a}^{2}+\mathrm{b}^{2}$ By taking LCM, we obtain: $\begin{array}{l} \frac{b^{2} x+a^{2} y}{a b}=a^{2}+b^{2} \\ \Rightarrow b^{2}...
Solve for x and y :
,
Solution: The given equations are: $\frac{b x}{a}-\frac{a y}{b}+a+b=0$ By taking LCM, we obtain: $b^{2} x-a^{2} y=-a^{2} b-b^{2} a\dots \dots(i)$ and $b x-a y+2 a b=0$ $b x-a y=-2 a b\dots...
Find the roots of the given equation:
$\begin{array}{l} 4 x^{2}+4 b x-\left(a^{2}-b^{2}\right)=0 \\ \Rightarrow 4 x^{2}+4 b x=a^{2}-b^{2} \\ \Rightarrow(2 x)^{2}+2 \times 2 x \times b+b^{2}=a^{2}-b^{2}+b^{2} \text { (Adding } b^{2}...
Solve for x and y:
,
Solution: The given equations are $\begin{array}{l} a x-b y=a^{2}+b^{2}\dots \dots(i) \\ x+y=2 a\dots \dots(ii) \end{array}$ From equation(ii) $y=2 a-x$ Substituting $\mathrm{y}=2...
Solve for x and y :
Solution: The given equations are $\begin{array}{l} 6(a x+b y)=3 a+2 b \\ \Rightarrow 6 a x+6 b y=3 a+2 b\dots \dots(i) \end{array}$ and $6(b x-a y)=3 b-2 a$ $\Rightarrow 6 b x-6 a y=3 b-2 a\dots...
Find the roots of the given equation:
$\frac{2}{x^{2}}-\frac{5}{x}+2=0$ $\Rightarrow \frac{2-5 x+2 x^{2}}{x^{2}}=0$ $\Rightarrow 2 x^{2}-5 x+2=0$ $\Rightarrow 4 x^{2}-10 x+4=0 \quad$ (Multiplying both sides by 2) $\Rightarrow 4 x^{2}-10...
Solve for x and y :
,
Solution: The given eq. can be written as $\begin{array}{l} \frac{x}{a}-\frac{y}{b}=0\dots \dots(i) \\ a x+b y-u^{2}+b^{2}\dots \dots(ii) \end{array}$ From equation(i), $\mathrm{y}=\frac{b x}{a}$...
Solve for x and y :
Solution: The given equations are $\begin{array}{l} \mathrm{px}+\mathrm{qy}=\mathrm{p}-\mathrm{q}\dots \dots(i) \\ \mathrm{qx}-\mathrm{py}=\mathrm{p}+\mathrm{q}\dots \dots(ii) \end{array}$...
Find the roots of the given equation:
$\begin{array}{l} 5 x^{2}-6 x-2=0 \\ \Rightarrow 25 x^{2}-30 x-10=0 \\ \Rightarrow 25 x^{2}-30 x=10 \end{array}$ $\begin{array}{l} \Rightarrow(5 x)^{2}-2 \times 5 x \times 3+3^{2}=10+3^{2} \\...
Solve for x and y :
,
Solution: The given equations are: $\begin{array}{l} \frac{x}{a}+\frac{y}{b}=2 \\ \Rightarrow \frac{b x+a y}{a b}=2 \quad[\text { Taking LCM }] \\ \Rightarrow \mathrm{bx}+\mathrm{ay}=2 \mathrm{ab}...
Find the roots of the given equation:
$3 x^{2}-2 x-1=0$ $\Rightarrow 9 x^{2}-6 x-3=0 \quad$ (Multiplying both sides by 3) $\Rightarrow 9 x^{2}-6 x=3$ $\Rightarrow(3 x)^{2}-2 \times 3 x \times 1+1^{2}=3+1^{2} \quad$ [Adding $1^{2}$ on...
Find the roots of the given equation:
$\begin{array}{l} 7 x^{2}+3 x-4=0 \\ \Rightarrow 49 x^{2}+21 x-28=0 \end{array}$ $\Rightarrow 49 x^{2}+21 x=28$ $\Rightarrow(7 x)^{2}+2 \times 7 x \times...
Find the roots of the given equation:
$8 x^{2}-14 x-15=0$ $\Rightarrow 16 x^{2}-28 x-30=0 \quad$ (Multiplying both sides by 2) $\Rightarrow 16 x^{2}-28 x=30$ $\Rightarrow(4 x)^{2}-2 \times 4 x \times...
Find the roots of the given equation:
$\begin{array}{l} 3 x^{2}-x-2=0 \\ \Rightarrow 9 x^{2}-3 x-6=0 \quad \text { (Multiplying both sides by 3) } \\ \Rightarrow 9 x^{2}-3 x=6 \\ \Rightarrow(3 x)^{2}-2 \times 3 x \times...
Find the roots of the given equation:
$2 x^{2}+5 x-3=0$ $\Rightarrow 4 x^{2}+10 x-6=0 \quad$ (Multiplying both sides by 2) $\Rightarrow 4 x^{2}+10 x=6$ $\Rightarrow(2 x)^{2}+2 \times 2 x \times...
Find the roots of the given equation:
$\begin{array}{l} 4 x^{2}+4 \sqrt{3} x+3=0 \\ \Rightarrow 4 x^{2}+4 \sqrt{3} x=-3 \\ \Rightarrow(2 x)^{2}+2 \times 2 x \times \sqrt{3}+(\sqrt{3})^{2}=-3+(\sqrt{3})^{2} \end{array}$ $\begin{array}{l}...
Find the roots of the given equation:
$\begin{array}{l} x^{2}+8 x-2=0 \\ \Rightarrow x^{2}+8 x=2 \\ \Rightarrow x^{2}+2 \times x \times 4+4^{2}=2+4^{2} \\ \Rightarrow(x+4)^{2}=2+16=18 \\ \Rightarrow x+4=\pm \sqrt{18}=\pm 3 \sqrt{2} \\...
Find the roots of the given equation:
$x^{2}-4 x+1=0$ $\Rightarrow x^{2}-4 x=-1$ $\Rightarrow x^{2}-2 \times x \times 2+2^{2}=-1+2^{2} \quad$ (Adding $2^{2}$ on both sides) $\Rightarrow(x-2)^{2}=-1+4=3$ $\Rightarrow x-2=\pm \sqrt{3}...
Find the roots of the given equation:
$\begin{array}{l} x^{2}-6 x+3=0 \\ \Rightarrow x^{2}-6 x=-3 \\ \Rightarrow x^{2}-2 \times x \times 3+3^{2}=-3+3^{2} \\ \Rightarrow(x-3)^{2}=-3+9=6 \\ \Rightarrow x-3=\pm \sqrt{6} \end{array}$...
Solve for x and y :
,
Solution: The given equations are $\begin{array}{l} \mathrm{x}+\mathrm{y}=\mathrm{a}+\mathrm{b}\dots \dots(i) \\ \mathrm{ax}-\mathrm{by}=\mathrm{a}^{2}-\mathrm{b}^{2}\dots \dots(ii) \end{array}$...
Solve for x and y :
,
Solution: The given equations can be written as $\begin{array}{l} \frac{3}{x}+\frac{6}{y}=7\dots \dots(i) \\ \frac{9}{x}+\frac{3}{y}=11\dots \dots(ii) \end{array}$ Multiplying equation(i) by 3 and...
Solve for x and y :
,
Solution: The given equations are $\begin{array}{l} \frac{2}{3 x+2 y}+\frac{3}{3 x-2 y}=\frac{17}{5}\dots \dots(i) \\ \frac{5}{3 x+2 y}+\frac{1}{3 x-2 y}=2\dots \dots(ii) \end{array}$ Substituting...
Solve for x and y :
,
where and
Solution: The given equations are $\begin{array}{l} \frac{1}{2(x+2 y)}+\frac{5}{3(3 x-2 y)}=-\frac{3}{2}\dots \dots(i) \\ \frac{1}{4(x+2 y)}-\frac{3}{5(3 x-2 y)}=\frac{61}{60}\dots \dots(ii)...
Solve for x and y:
,
Solution: The given equations are $\begin{array}{l} \frac{1}{3 x+y}+\frac{1}{3 x-y}=\frac{3}{4}\dots \dots(i) \\ \frac{1}{2(3 x+y)}-\frac{1}{2(3 x-y)}=-\frac{1}{8} \end{array}$ $\frac{1}{3...
Solve for x and y :
,
Solution: The given equations can be written as $\begin{array}{l} \frac{5}{x}+\frac{2}{y}=6\dots \dots (i) \\ \frac{-5}{x}+\frac{4}{y}=-3\dots \dots(ii) \end{array}$ Adding equation(i) and...
Solve for x and y : ,
Solution: The given equations are: $23 x-29 y=98 \quad \ldots .$ (i) $29 x-23 y=110 \ldots \ldots$ (ii) Adding equation(i) and equation(ii), we obtain: $\begin{array}{l} 52 x-52 y=208 \\ \Rightarrow...
Solve for x and y:
217 x+131 y=913,
131 x+217 y=827
Solution: The given equations are: $\begin{aligned} 217 \mathrm{x}+131 \mathrm{y} =913\dots \dots(i) \\ 131 \mathrm{x}+217 \mathrm{y} =827 \ldots \text {....(ii) } \end{aligned}$ On adding...
Solve for x and y :
,
Solution: The given eq. are: $\begin{array}{l} 71 x+37 y=253\dots \dots(i) \\ 37 x+71 y=287 \ldots \ldots \text {...(ii) } \end{array}$ On adding equation(i) and equation(ii), we obtain:...
Solve for x and y :
,
, where
Solution: The given equations are $\frac{10}{x+y}+\frac{2}{x-y}=4 \quad \ldots \ldots$ (i) $\frac{15}{x+y}-\frac{9}{x-y}=-2\dots \dots(ii)$ Substituting $\frac{1}{x+y}=\mathrm{u}$ and...
Solve for x and y :
,
Solution: The given eq. are $\frac{44}{x+y}+\frac{30}{x-y}=10 \dots \dots(i)$ $\frac{55}{x+y}-\frac{40}{x-y}=13\dots \dots (ii)$ Putting $\frac{1}{x+y}=u$ and $\frac{1}{x-y}=v$, we get: $44 u+30...
Solve for x and y : , , where
Solution: The given eq. are $\frac{5}{x+1}+\frac{2}{y-1}=\frac{1}{2} \quad \ldots \ldots$ (i) $\frac{10}{x+1}-\frac{2}{y-1}=\frac{5}{2}\dots \dots (ii)$ On substituting $\frac{1}{x+1}=\mathrm{u}$...
Solve for x and y :
,
Solution: The given eq. are $\begin{array}{l} \frac{3}{x+y}+\frac{2}{x-y}=2 \ldots \ldots \text { (i) } \\ \frac{9}{x+y}-\frac{4}{x-y}=1 \ldots \ldots \text { (ii) } \end{array}$ Substituting...
Solve for x and y :
,
Solution: The given eq. are $\frac{5}{x+y}-\frac{2}{x-y}=-1 \quad \ldots \ldots \text { (i) }$ $\frac{15}{x+y}-\frac{7}{x-y}=10\dots \dots(ii)$ Substituting $\frac{1}{x+y}=\mathrm{u}$ and...
Solve for x and y:
x + y = 5xy,
3x + 2y = 13xy
Solution: The given eq. are: $x + y = 5xy \dots \dots(i)$ $3x + 2y = 13xy \dots \dots(ii)$ From equation (i), we have: $\begin{array}{l} \frac{x+y}{x y}=5 \\ \Rightarrow \frac{1}{y}+\frac{1}{x}=5...
Solve for x and y :
,
Solution: The given eq. are: $\begin{array}{l} 4 x+6 y=3 x y \quad \ldots \ldots(i) \\ 8 x+9 y=5 x y \quad \ldots \ldots(i i) \end{array}$ From eq. (i), we have: $\begin{array}{l} \frac{4 x+6 y}{x...
Solve for x and y :
,
Solution: The given eq. are: $\begin{array}{l} \frac{3}{x}+\frac{2}{y}=12 \ldots \ldots \ldots \text { (i) } \\ \frac{2}{x}+\frac{3}{y}=13 \ldots \ldots . \text { (ii) } \end{array}$ Multiplying...
Solve for x and y :
,
Solution: The given eq. are: $\begin{array}{l} \frac{5}{x}-\frac{3}{y}=1 \ldots \ldots(i) \\ \frac{3}{2 x}+\frac{2}{3 y}=5 \ldots \ldots (ii) \end{array}$ Putting $\frac{1}{x}=u$ and...
Solve for x and y:
,
Solution: The given eq. are: $\begin{array}{l} \frac{9}{x}-\frac{4}{y}=8 \ldots \ldots . .(i) \\ \frac{13}{x}+\frac{7}{y}=101 \ldots \ldots (ii) \end{array}$ Putting $\frac{1}{x}=u$ and...
Find the roots of the given equation:
$\begin{array}{l} 2^{2 x}-3.2^{(x+2)}+32=0 \\ \Rightarrow\left(2^{x}\right)^{2}-3.2^{x} \cdot 2^{2}+32=0 \end{array}$ Let $2^{x}$ be $y$. $\begin{array}{l} \therefore y^{2}-12 y+32=0 \\ \Rightarrow...
Find the roots of the given equation:
Given: $\begin{array}{l} 4^{(x+1)}+4^{(1-x)}=10 \\ \Rightarrow 4^{x} \cdot 4+4^{1} \cdot \frac{1}{4^{4}}=10 \end{array}$ Let $4^{x}$ be $y$. $\begin{array}{l} \therefore 4 y+\frac{4}{y}=10 \\...
Find the roots of the given equation:
$\begin{array}{l} 3^{(x+2)}+3^{-x}=10 \\ 3^{x} .9+\frac{1}{3^{x}}=10 \end{array}$ Let $3^{x}$ be equal to $y$. $\begin{array}{l} \therefore 9 y+\frac{1}{y}=10 \\ \Rightarrow 9 y^{2}+1=10 y \\...
Find the roots of the given equation:
$\begin{array}{l} \frac{a}{(a x-1)}+\frac{b}{(b x-1)}=(a+b) \\ \Rightarrow\left[\frac{a}{(a x-1)}-b\right]+\left[\frac{b}{(b x-1)}-a\right]=0 \\ \Rightarrow \frac{a-b(a x-1)}{a x-1}+\frac{b-a(b...
Find the roots of the given equation:
$\begin{array}{l} \frac{a}{(x-b)}+\frac{b}{(x-a)}=2 \\ \Rightarrow\left[\frac{a}{(x-b)}-1\right]+\left[\frac{b}{(x-b)}-1\right]=0 \\ \Rightarrow \frac{a-(x-b)}{x-b}+\frac{b-(x-b)}{x-b}=0...
Find the roots of the given equation:
$\left(\frac{x}{x+1}\right)^{2}-5\left(\frac{x}{x+1}\right)+6=0$ Putting $\frac{x}{x+1}=y$, we get: $\begin{array}{l} y^{2}-5 y+6=0 \\ \Rightarrow y^{2}-5 y+6=0 \end{array}$ $\begin{array}{l}...
Find the roots of the given equation:
Given: $\left(\frac{4 x-3}{2 x+1}\right)-10\left(\frac{2 x+1}{4 x-3}\right)=3$ Putting $\frac{4 x-3}{2 x+1}=y$, we get: $\begin{array}{l} y-\frac{10}{y}=3 \\ \Rightarrow \frac{y^{2}-10}{y}=3 \\...
Find the roots of the given equation:
$\begin{array}{l} 3\left(\frac{7 x+1}{5 x-3}\right)-4\left(\frac{5 x-3}{7 x+1}\right)=11, x \neq \frac{3}{5},-\frac{1}{7} \\ \Rightarrow \frac{3(7 x+1)^{2}-4(5 x-3)^{2}}{(5 x-3)(7 x+1)}=11 \\...
Find the roots of the given equation:
$\begin{array}{l} 3\left(\frac{3 x-1}{2 x+3}\right)-2\left(\frac{2 x+3}{3 x-1}\right)=5, x \neq \frac{1}{3},-\frac{3}{2} \\ \Rightarrow \frac{3(3 x-1)^{2}-2(2 x+3)^{2}}{(2 x+3)(3 x-1)}=5 \\...
Find the roots of the given equation:
$\begin{array}{l} \frac{1}{x+1}+\frac{2}{x+2}=\frac{5}{x+4}, x \neq-1,-2,-4 \\ \Rightarrow \frac{x+2+2 x+2}{(x+1)(x+2)}=\frac{5}{x+4} \\ \Rightarrow \frac{3 x+4}{x^{2}+3 x+2}=\frac{5}{x+4} \\...
Find the roots of the given equation:
$\begin{array}{l} \frac{1}{(x-2)}+\frac{2}{(x-1)}=\frac{6}{x} \\ \Rightarrow \frac{(x-1)+2(x-2)}{(x-1)(x-2)}=\frac{6}{x} \\ \Rightarrow \frac{3 x-5}{x^{2}-3 x+2}=\frac{6}{x} \\ \Rightarrow 3 x^{2}-5...
Find the roots of the given equation:
$\begin{array}{l} \frac{x-1}{x-2}+\frac{x-3}{x-4}=3 \frac{1}{3}, x \neq 2,4 \\ \Rightarrow \frac{(x-1)(x-4)+(x-2)(x-3)}{(x-2)(x-4)}=\frac{10}{3} \\ \Rightarrow \frac{x^{2}-5 x+4+x^{2}-5 x+6}{x^{2}-6...
Solve for x and y :
,
Solution: The given eq. are: $\begin{array}{l} \frac{3}{x}-\frac{1}{y}+9=0 \\ \Rightarrow \frac{3}{x}-\frac{1}{y}=-9 \quad \ldots \ldots(i) \\ \Rightarrow \frac{2}{x}-\frac{3}{y}=5 \ldots \ldots...
Find the roots of the given equation:
$\begin{array}{l} \frac{x-4}{x-5}+\frac{x-6}{x-7}=3 \frac{1}{3}, x \neq 5,7 \\ \Rightarrow \frac{(x-4)(x-7)+(x-5)(x-6)}{(x-5)(x-7)}=\frac{10}{3} \\ \Rightarrow \frac{x^{2}-11 x+28+x^{2}-11...
Find the roots of the given equation:
$\begin{array}{l} \frac{x}{x+1}+\frac{x+1}{x}=2 \frac{4}{15}, x \neq 0,-1 \\ \Rightarrow \frac{x^{2}+(x+1)^{2}}{x(x+1)}=\frac{34}{15} \\ \Rightarrow \frac{x^{2}+x^{2}+2 x+1}{x^{2}+x}=\frac{34}{15}...
Find the roots of the given equation:
$\begin{array}{l} \frac{x}{x-1}+\frac{x-1}{x}=4 \frac{1}{4}, x \neq 0,1 \\ \Rightarrow \frac{x^{2}+(x-1)^{2}}{x(x-1)}=\frac{17}{4} \\ \Rightarrow \frac{x^{2}+x^{2}-2 x+1}{x^{3}-x}=\frac{17}{4} \\...
Find the roots of the given equation:
$\begin{array}{l} \frac{3 x-4}{7}+\frac{7}{3 x-4}=\frac{5}{2}, x \neq \frac{4}{3} \\ \Rightarrow \frac{(3 x-4)^{2}+49}{7(3 x-4)}=\frac{5}{2} \end{array}$ $\begin{array}{l} \Rightarrow \frac{9...
Find the roots of the given equation: 57.
$\begin{array}{l} \frac{(x+3)}{(x-2)}-\frac{(1-x)}{x}=\frac{17}{4} \\ \Rightarrow \frac{x(x+3)-(1-x)(x-2)}{(x-2) x}=\frac{17}{4} \\ \Rightarrow \frac{x^{2}+3 x-\left(x-2-x^{2}+2 x\right)}{x^{2}-2...
Find the roots of the given equation:
$\begin{array}{l} \frac{1}{2 a+b+2 x}=\frac{1}{2 a}+\frac{1}{b}+\frac{1}{2 x} \\ \Rightarrow \frac{1}{2 a+b+2 x}-\frac{1}{2 x}=\frac{1}{2 a}+\frac{1}{b} \\ \Rightarrow \frac{2 x-2 a-b-2 x}{2 x(2...
Solve for x and y:
,
Solution: The given eq. are: $\begin{array}{l} 2 x-\frac{3}{y}=9 \ldots \ldots \text { (i) } \\ 3 x+\frac{7}{y}=2 \ldots \ldots \text { (ii) } \end{array}$ Putting $\frac{1}{y}=\mathrm{v}$, we...
Find the roots of the given equation:
$\begin{array}{l} \frac{1}{x-1}-\frac{1}{x+5}=\frac{6}{7}, x \neq 1,-5 \\ \Rightarrow \frac{x+5-x+1}{(x-1)(x+5)}=\frac{6}{7} \\ \Rightarrow \frac{6}{x^{2}+4 x-5}=\frac{6}{7} \\ \Rightarrow x^{2}+4...
Find the roots of the given equation:
$\begin{array}{l} \frac{3}{x+1}-\frac{1}{2}=\frac{2}{3 x-1}, x \neq-1, \frac{1}{3} \\ \Rightarrow \frac{3}{x+1}-\frac{2}{3 x-1}=\frac{1}{2} \\ \Rightarrow \frac{9 x-3-2 x-2}{(x+1)(3...
Find the roots of the given equation:
$\begin{array}{l} \frac{4}{x}-3=\frac{5}{2 x+3}, x \neq 0,-\frac{3}{2} \\ \Rightarrow \frac{4}{x}-\frac{5}{2 x+3}=3 \\ \Rightarrow \frac{8 x+12-5 x}{x(2 x+3)}=3 \\ \Rightarrow \frac{3 x+12}{2...
Find the roots of the given equation:
$\begin{array}{l} \frac{16}{x}-1=\frac{15}{x+1}, x \neq 0,-1 \\ \Rightarrow \frac{16}{x}-\frac{15}{x+1}=1 \\ \Rightarrow \frac{16 x+16-15 x}{x(x+1)}=1 \\ \Rightarrow \frac{x+16}{x^{2}+x}=1 \\...
Solve for x and y :
,
Solution: The given eq. are: $\begin{array}{l} x+\frac{6}{y}=6 \quad \ldots \ldots \text {(i) } \\ 3 x-\frac{8}{y}=5 \ldots \ldots \text { (ii) } \end{array}$ Putting $\frac{1}{y}=\mathrm{v}$, we...
Find the roots of the given equation:
We write, $-9(a+b) x=-3(2 a+b) x-3(a+2 b) x$ as $\begin{array}{l} 9 x^{2} \times\left(2 a^{2}+5 a b+2 b^{2}\right)=9\left(2 a^{2}+5 a b+2 b^{2}\right) x^{2}=[-3(2 a+b) x] \times[-3(a+2 b) x] \\...
Solve for x and y :
Solution: The given eq. are: $\begin{array}{l} \frac{5}{x}+6 y=13 \ldots \ldots \text { (i) } \\ \frac{3}{x}+4 y=7 \ldots \ldots \text { (ii) } \end{array}$ Putting $\frac{1}{x}=\mathrm{u}$, we...
Find the roots of the given equation:
$\begin{array}{l} a^{2} b^{2} x^{2}+b^{2} x-a^{2} x-1=0 \\ \Rightarrow b^{2} x\left(a^{2} x+1\right)-1\left(a^{2} x+1\right)=0 \\ \Rightarrow\left(b^{2} x-1\right)\left(a^{2} x+1\right)=0 \\...
Find the roots of the given equation:
$\begin{array}{l} 12 a b x^{2}-\left(9 a^{2}-8 b^{2}\right) x-6 a b=0 \\ \Rightarrow 12 a b x^{2}-9 a^{2} x+8 b^{2} x-6 a b=0 \\ \Rightarrow 3 a x(4 b x-3 a)+2 b(4 b x-3 a)=0 \\ \Rightarrow(3 a x+2...
Find the roots of the given equation:
$\begin{array}{l} 4 x^{2}-2\left(a^{2}+b^{2}\right) x+a^{2} b^{2}=0 \\ \Rightarrow 4 x^{2}-2 a^{2} x-2 b^{2} x+a^{2} b^{2}=0 \\ \Rightarrow 2 x\left(2 x-a^{2}\right)-b^{2}\left(2 x-a^{2}\right)=0 \\...
Solve for x and y :
Solution: The given eq. are: $\frac{x+y-8}{2}=\frac{x+2 y-14}{3}=\frac{3 x+y-12}{11}$ i.e., $\frac{x+y-8}{2}=\frac{3 x+y-12}{11}$ On cross multiplication, we obtain: $\begin{array}{l} 11 x+11 y-88=6...
Find the roots of the given equation:
We write, $-4 a x=-(b+2 a) x+(b-2 a) x$ as $\begin{array}{l} x^{2} \times\left(-b^{2}+4 a^{2}\right)=\left(-b^{2}+4 a^{2}\right) x^{2}=-(b+2 a) x \times(b-2 a) x \\ \therefore x^{2}-4 a x-b^{2}+4...
Find the roots of the given equation:
$\begin{array}{l} a b x^{2}+\left(b^{2}-a c\right) x-b c=0 \\ \Rightarrow a b x^{2}+b^{2} x-a c x-b c=0 \\ \Rightarrow b x(a x+b)-c(a x+b)=0 \\ \Rightarrow(b x-c)(a x+b)=0 \\ \Rightarrow b x-c=0...
Find the roots of the given equation:
We write, $6 x=(a+4) x-(a-2) x$ as $\begin{array}{l} x^{2} \times\left[-\left(a^{2}+2 a-8\right)\right]=-\left(a^{2}+2 a-8\right) x^{2}=(a+4) x \times[-(a-2) x] \\ \therefore x^{2}+6 x-\left(a^{2}+2...
Solve for x and y:
6x + 5y = 7x + 3y + 1 = 2(x + 6y – 1)
Solution: The given equations are: $\begin{array}{l} 6 x+5 y=7 x+3 y+1=2(x+6 y-1) \\ \Rightarrow 6 x+5 y=2(x+6 y-1) \\ \Rightarrow 6 x+5 y=2 x+12 y-2 \\ \Rightarrow 6 x-2 x+5 y-12 y=-2 \\...
Find the roots of the given equation:
We write, $-(2 b-1) x=-(b-5) x-(b+4) x$ as $\begin{array}{l} x^{2} \times\left(b^{2}-b-20\right)=\left(b^{2}-b-20\right) x^{2}=[-(b-5) x] \times[-(b+4) x] \\ \therefore x^{2}-(2 b-1)...
Find the roots of the given equation:
We have, $-2 a x=(2 b-a) x-(2 b+a) x$ as $\begin{array}{l} x^{2} \times\left[-\left(4 b^{2}-a^{2}\right)\right]=-\left(4 b^{2}-a^{2}\right) x^{2}=(2 b-a) x \times[-(2 b+a) x] \\ \therefore x^{2}-2 a...
Find the roots of the given equation:
We write, $5 x=(a+3) x-(a-2) x$ as $\begin{array}{l} x^{2} \times\left[-\left(a^{2}+a-6\right)\right]=-\left(a^{2}+a-6\right) x^{2}=(a+3) x \times[-(a-2) x] \\ \therefore x^{2}+5...
Find the roots of the given equation:
We write, $-4 a^{2} x=-2\left(a^{2}+b^{2}\right) x-2\left(a^{2}-b^{2}\right) x$ as $\begin{array}{l} 4 x^{2} \times\left(a^{4}-b^{4}\right)=4\left(a^{4}-b^{4}\right)...
Find the roots of the given equation:
We write, $4 b x=2(a+b) x-2(a-b) x$ as $\begin{array}{l} 4 x^{2} \times\left[-\left(a^{2}-b^{2}\right)\right]=-4\left(a^{2}-b^{2}\right) x^{2}=2(a+b) x \times[-2(a-b) x] \\ \therefore 4 x^{2}+4 b...
Solve for x and y:
7(y + 3) – 2(x + 2) = 14,
4(y – 2) + 3(x – 3) = 2
Solution: The given eq. are: $7(y + 3) – 2(x + 2) = 14$ $\Rightarrow 7y + 21 – 2x – 4 = 14$ $\Rightarrow -2x + 7y = -3 \dots \dots(i)$ and $4(y – 2) + 3(x – 3) = 2$ $\Rightarrow 4y – 8 + 3x – 9 = 2$...
Find the roots of the given equation:
We write, $a x=2 a x-a x$ as $2 x^{2} \times\left(-a^{2}\right)=-2 a^{2} x^{2}=2 a x \times(-a x)$ $\begin{array}{l} \therefore 2 x^{2}+a x-a^{2}=0 \\ \Rightarrow 2 x^{2}+2 a x-a x-a^{2}=0 \\...
Find the roots of the given equation:
$\begin{array}{l} \frac{2}{x^{2}}-\frac{5}{x}+2=0 \\ \Rightarrow 2-5 x+2 x^{2}=0 \\ \Rightarrow 2 x^{2}-5 x+2=0 \\ \Rightarrow 2 x^{2}-(4 x+x)+2=0 \\ \Rightarrow 2 x^{2}-4 x-x+2=0 \\ \Rightarrow 2...
Find the roots of the given equation:
$\begin{array}{l} 10 x-\frac{1}{x}=3 \\ \Rightarrow 10 x^{2}-1=3 x \end{array}$ [Multiplying both sides by $x]$ $\begin{array}{l} \Rightarrow 10 x^{2}-3 x-1=0 \\ \Rightarrow 10 x^{2}-(5 x-2 x)-1=0...
Find the roots of the given equation:
We write, $-x=-\frac{x}{2}-\frac{x}{2}$ as $2 x^{2} \times \frac{1}{8}=\frac{x^{2}}{4}=\left(-\frac{x}{2}\right) \times\left(-\frac{x}{2}\right)$ $\begin{array}{l} \therefore 2 x^{2}-x+\frac{1}{8}=0...
Solve for x and y:
0.3x + 0.5y = 0.5,
0.5x + 0.7y = 0.74
Solution: The given system of eq. is $0.3 \mathrm{x}+0.5 \mathrm{y}=0.5\dots \dots(i)$ $0.5 \mathrm{x}+0.7 \mathrm{y}=0.74 \dots \dots(ii)$ On multiplying equation(i) by 5 and equation(ii) by 3 and...
Find the roots of the given equation:
We write, $-20 x=-10 x-10 x$ as $100 x^{2} \times 1=100 x^{2}=(-10 x) \times(-10 x)$ $\begin{array}{l} \therefore 100 x^{2}-20 x+1=0 \\ \Rightarrow 100 x^{2}-10 x-10 x+1=0 \end{array}$...
Find the roots of the given equation:
$\begin{array}{l} 9 x^{2}+6 x+1=0 \\ \Rightarrow 9 x^{2}+3 x+3 x+1=0 \\ \Rightarrow 3 x(3 x+1)+1(3 x+1)=0 \\ \Rightarrow(3 x+1)(3 x+1)=0 \\ \Rightarrow 3 x+1=0 \text { or } 3 x+1=0 \\ \Rightarrow...
Find the roots of the given equation:
$\begin{array}{l} x^{2}-(1+\sqrt{2}) x+\sqrt{2}=0 \\ \Rightarrow x^{2}-x-\sqrt{2} x+\sqrt{2}=0 \\ \Rightarrow x(x-1)-\sqrt{2}(x-1)=0 \\ \Rightarrow(x-\sqrt{2})(x-1)=0 \\ \Rightarrow x-\sqrt{2}=0...
Solve for x and y:
0.4x + 0.3y = 1.7,
0.7x – 0.2y = 0.8.
Solution: The given system of eq. is $0.4x + 0.3y = 1.7 \dots \dots(i)$ $0.7x – 0.2y = 0.8 \dots \dots(ii)$ On multiplying equation(i) by $0.2$ and equation(ii) by $0.3$ and adding them, we obtain...
Find the roots of the given equation:
We write, $13 x=5 x+8 x$ as $5 x^{2} \times 8=40 x^{2}=5 x \times 8 x$ $\begin{array}{l} \therefore 5 x^{2}+13 x+8=0 \\ \Rightarrow 5 x^{2}+5 x+8 x+8=0 \\ \Rightarrow 5 x(x+1)+8(x+1)=0 \\...
Find the roots of the given equation:
We write, $7 x=5 x+2 x$ as $\sqrt{2} x^{2} \times 5 \sqrt{2}=10 x^{2}=5 x \times 2 x$ $\begin{array}{l} \therefore \sqrt{2} x^{2}+7 x+5 \sqrt{2}=0 \\ \Rightarrow \sqrt{2} x^{2}+5 x+2 x+5 \sqrt{2}=0...
Solve for x and y :
Solution: The given eq. are: $\begin{array}{l} \frac{7-4 x}{3}=y \\ \Rightarrow 4 x+3 y=7 \dots \dots(i) \end{array}$ $\begin{array}{l} \text { and } 2 x+3 y+1=0 \\ \Rightarrow 2 x+3 y=-1 \ldots...
Find the roots of the given equation:
We write, $3 \sqrt{3} x=5 \sqrt{3} x-2 \sqrt{3} x$ as $x^{2} \times(-30)=-30 x^{2}=5 \sqrt{3} x \times(-2 \sqrt{3} x)$ $\begin{array}{l} \therefore x^{2}+3 \sqrt{3} x-30=0 \\ \Rightarrow x^{2}+5...
Find the roots of the given equation:
$\begin{array}{l} x^{2}-(\sqrt{3}+1) x+\sqrt{3}=0 \\ \Rightarrow x^{2}-\sqrt{3} x-x+\sqrt{3}=0 \\ \Rightarrow x(x-\sqrt{3})-1(x-\sqrt{3})=0 \\ \Rightarrow(x-\sqrt{3})(x-1)=0 \\ \Rightarrow...
Find the roots of the given equation:
We write, $-3 \sqrt{5} x=-2 \sqrt{5} x-\sqrt{5} x$ as $x^{2} \times 10=10 x^{2}=(-2 \sqrt{5} x) \times(-\sqrt{5} x)$ $\begin{array}{l} \therefore x^{2}-3 \sqrt{5} x+10=0 \\ \Rightarrow x^{2}-2...
Find the roots of the given equation:
We write, $-2 \sqrt{2} x=-3 \sqrt{2} x+\sqrt{2} x$ as $\sqrt{3} x^{2} \times(-2 \sqrt{3})=-6 x^{2}=(-3 \sqrt{2} x) \times(\sqrt{2} x)$ $\therefore \sqrt{3} x^{2}-2 \sqrt{2} x-2 \sqrt{3}=0$...
Find the roots of the given equation:
We write, $-2 \sqrt{6} x=-\sqrt{6} x$ ad $3 x^{2} \times 2=6 x^{2}=(-\sqrt{6} x) \times(-\sqrt{6} x)$ $\begin{array}{l} \therefore 3 x^{2}-2 \sqrt{6} x+2=0 \\ \Rightarrow 3 x^{2}-\sqrt{6} x-\sqrt{6}...
If one zero of the quadratic polynomial is 2 , then the value of is (a) (b) (c) (d)
The correct option is option (d) $\frac{-6}{5}$ Since 2 is a zero of $k x^{2}+3 x+k$, we have: $\mathrm{k} \times(2)^{2}+3(2)+\mathrm{k}=0$ $\Rightarrow 4 \mathrm{k}+\mathrm{k}+6=0$ $\Rightarrow 5...
Find the roots of the given equation:
$\begin{array}{l} 4 \sqrt{6} x^{2}-13 x-2 \sqrt{6}=0 \\ \Rightarrow 4 \sqrt{6} x^{2}-16 x+3 x-2 \sqrt{6}=0 \\ \Rightarrow 4 \sqrt{2} x(\sqrt{3} x-2 \sqrt{2})+\sqrt{3}(\sqrt{3} x-2 \sqrt{2})=0 \\...
Find the roots of the given equation:
We write, $-6 x=7 x-13 x$ as $\sqrt{7} x^{2} \times(-13 \sqrt{7})=-91 x^{2}=7 x \times(-13 x)$ $\begin{array}{l} \therefore \sqrt{7} x^{2}-6 x-13 \sqrt{7}=0 \\ \Rightarrow \sqrt{7} x^{2}+7 x-13 x-13...
Solve for x and y :
,
Solution: The given eq. are; $\begin{array}{l} 2 x-5 y=\frac{8}{3} \ldots \ldots(i) \\ 3 x-2 y=\frac{5}{6} \ldots \cdots . . \text { (ii) } \end{array}$ On multiplying equation(i) by 2 and...
Find the roots of the given equation:
$\begin{array}{l} 3 \sqrt{7} x^{2}+4 x-\sqrt{7}=0 \\ \Rightarrow 3 \sqrt{7} x^{2}+7 x-3 x-\sqrt{7}=0 \\ \Rightarrow \sqrt{7} x(3 x+\sqrt{7})-1(3 x+\sqrt{7})=0 \\ \Rightarrow(3...
Find the roots of the given equation:
$\begin{array}{l} \sqrt{3} x^{2}+11 x+6 \sqrt{3}=0 \\ \Rightarrow \sqrt{3} x^{2}+9 x+2 x+6 \sqrt{3}=0 \\ \Rightarrow \sqrt{3} x(x+3 \sqrt{3})+2(x+3 \sqrt{3})=0 \end{array}$ $\begin{array}{l}...
Find the roots of the given equation:
We write: $10 x=3 x+7 x$ as $\sqrt{3} x^{2} \times 7 \sqrt{3}=21 x^{2}=3 x \times 7 x$ $\begin{array}{l} \therefore \sqrt{3} x^{2}+10 x+7 \sqrt{3}=0 \\ \Rightarrow \sqrt{3} x^{2}+3 x+7 x+7...
Find the roots of the given equation:
We write: $2 \sqrt{2} x=3 \sqrt{2} x-\sqrt{2} x$ as $x^{2} \times(-6)=-6 x^{2}=3 \sqrt{2} x \times(-\sqrt{2} x)$ $\therefore x^{2}+2 \sqrt{2} x-6=0$ $\Rightarrow x^{2}+2 \sqrt{2} x-\sqrt{2} x-6=0$...
Find the roots of the given equation:
$\begin{array}{l} 48 x^{2}-13 x-1=0 \\ \Rightarrow 48 x^{2}-(16 x-3 x)-1=0 \\ \Rightarrow 48 x^{2}-16 x+3 x-1=0 \\ \Rightarrow 16 x(3 x-1)+1(3 x-1)=0 \\ \Rightarrow(16 x+1)(3 x-1)=0 \\ \Rightarrow...
Find the roots of the given equation:
$\begin{array}{l} 4-11 x=3 x^{2} \\ \Rightarrow 3 x^{2}+11 x-4=0 \\ \Rightarrow 3 x^{2}+12 x-x-4=0 \\ \Rightarrow 3 x(x+4)-1(x+4)=0 \\ \Rightarrow(x+4)(3 x-1)=0 \\ \Rightarrow x+4=0 \text { or } 3...
Solve for x and y :
,
Solution: The given eq. are: $\begin{array}{l} 2 \mathrm{x}-\frac{3 y}{4}=3 \ldots \ldots(\mathrm{i}) \\ 5 \mathrm{x}=2 \mathrm{y}+7 \ldots \ldots \text { (ii) } \end{array}$ When multiplying...
Find the roots of the given equation:
$\begin{array}{l} 15 x^{2}-28=x \\ \Rightarrow 15 x^{2}-x-28=0 \\ \Rightarrow 15 x^{2}-(21 x-20 x)-28=0 \\ \Rightarrow 15 x^{2}-21 x+20 x-28=0 \end{array}$ $\begin{array}{l} \Rightarrow 3 x(5...
Find the roots of the given equation:
$\begin{array}{l} 4 x^{2}-9 x=100 \\ \Rightarrow 4 x^{2}-9 x-100=0 \\ \Rightarrow 4 x^{2}-(25 x-16 x)-100=0 \\ \Rightarrow 4 x^{2}-25 x+16 x-100=0 \\ \Rightarrow x(4 x-25)+4(4 x-25)=0 \\...
Solve for x and y :
Solution: The given system of eq. is: $\begin{array}{ll} 4 \mathrm{x}-3 \mathrm{y}=8 & \ldots \ldots \text { (i) } \\ 6 \mathrm{x}-\mathrm{y}=\frac{29}{3} & \ldots \ldots \text { (ii) }...
Find the roots of the given equation:
We write, $-2 x=-3 x+x$ as $3 x^{2} \times(-1)=-3 x^{2}=(-3 x) \times x$ $\begin{array}{l} \therefore 3 x^{2}-2 x-1=0 \\ \Rightarrow 3 x^{2}-3 x+x-1=0 \\ \Rightarrow 3 x(x-1)+1(x-1)=0 \\...
If the zeroes of the polynomial are , a and , find the values of a and b.
using the relationship between the zeroes of he quadratic polynomial. Sum of zeroes $=\frac{-\left(\text { coef ficient of } x^{2}\right)}{\text { coefficient of } x^{3}}$ $\therefore...
Find the roots of the given equation: .
$\begin{array}{l} 6 x^{2}+x-12=0 \\ \Rightarrow 6 x^{2}+9 x-8 x-12=0 \\ \Rightarrow 3 x(2 x+3)-4(2 x+3)=0 \\ \Rightarrow(3 x-4)(2 x+3)=0 \\ \Rightarrow 3 x-4=0 \text { or } 2 x+3=0 \\ \Rightarrow...
Find the roots of the given equation: .
$\begin{array}{l} 6 x^{2}+11 x+3=0 \\ \Rightarrow 6 x^{2}+9 x+2 x+3=0 \\ \Rightarrow 3 x(2 x+3)+1(2 x+3)=0 \\ \Rightarrow(3 x+1)(2 x+3)=0 \\ \Rightarrow 3 x+1=0 \text { or } 2 x+3=0 \\ \Rightarrow...
Solve for and :
Solution: The given eq. are: $\begin{array}{l} \frac{x}{3}+\frac{y}{4}=11 \\ \Rightarrow 4 x+3 y=132 \ldots \ldots(i) \end{array}$ and $\frac{5 x}{6}-\frac{y}{3}+7=0$ $\Rightarrow 5 x-2 y=-42 \ldots...
If are the zeroes of the polynomial , then .
using the relationship between the zeroes of the quadratic polynomial. We have Sum of zeroes $=\frac{-(\text { coef ficient of } x)}{\text { coefficient of } x^{2}}$ and Product of zeroes...
If are the zeroes of the polynomial , then
using the relationship between the zeroes of he quadratic polynomial. => Sum of zeroes $=\frac{-\text { (coef ficient of } x)}{\text { coefficient of } x^{2}}$ and Product of zeroes $=\frac{\text...
Find the roots of the given equation:
$\begin{array}{l} x^{2}=18 x-77 \\ \Rightarrow x^{2}-18 x+77=0 \end{array}$ $\begin{array}{l} \Rightarrow x^{2}-(11 x+7 x)+77=0 \\ \Rightarrow x^{2}-11 x-7 x+77=0 \\ \Rightarrow x(x-11)-7(x-11)=0 \\...
Find the roots of the given equation:
$\begin{array}{l} x^{2}+12 x+35=0 \\ \Rightarrow x^{2}+7 x+5 x+35=0 \\ \Rightarrow x(x+7)+5(x+7)=0 \\ \Rightarrow(x+5)(x+7)=0 \\ \Rightarrow x+5=0 \text { or } x+7=0 \\ \Rightarrow x=-5 \text { or }...
If and are the zeros of the polynomial find the value of
using the relationship between the zeroes of the quadratic polynomial. Sum of zeroes $=\frac{-(\text { coef ficient of } x)}{\text { coefficient of } x^{2}}$ and Product of zeroes $=\frac{\text {...
Solve for x and y:
9x – 2y = 108,
3x + 7y = 105
Solution: The given system of eq. can be written as: $\begin{array}{cc} 9 \mathrm{x}-2 \mathrm{y}=108 & \ldots \ldots \text { (i) } \\ 3 \mathrm{x}+7 \mathrm{y}=105 & \ldots \text {..(ii) }...
Find the roots of the given equation: .
We write, $-3 x=3 x-6 x$ as $9 x^{2} \times(-2)=-18 x^{2}=3 x \times(-6 x)$ $\begin{array}{l} \therefore 9 x^{2}-3 x-2=0 \\ \Rightarrow 9 x^{2}+3 x-6 x-2=0 \\ \Rightarrow 3 x(3 x+1)-2(3 x+1)=0 \\...
If are the zeroes of the polynomial such that , find the value of
using the relationship between the zeroes of the quadratic polynomial. Sum of zeroes $=\frac{-(\text { coef ficient of } x)}{\text { coefficient of } x^{2}}$ and Product of zeroes $=\frac{\text {...
Find the roots of the given equation:
We write, $6 x=x+5 x$ as $x^{2} \times 5=5 x^{2}=x \times 5 x$ $\begin{array}{l} \therefore x^{2}+6 x+5=0 \\ \Rightarrow x^{2}+x-5 x+5=0 \end{array}$ $\begin{array}{l} \Rightarrow x(x+1)+5(x+1)=0 \\...
Find the zeroes of the quadratic polynomial .
For finding the zeroes of the quadratic polynomial we will equate $f(x)$ to 0 Hence, the zeroes of the quadratic polynomial $f(x)=4 \sqrt{3} x^{2}+5 x-2 \sqrt{3}$ are $-\frac{2}{\sqrt{3}}$ or...
Solve for x and y:
2x – y + 3 = 0,
3x – 7y + 10 = 0
Solution: The given system of eq. is: $\begin{array}{l} 2 x-y+3=0 \ldots \ldots(i) \\ 3 x-7 y+10=0 \ldots \ldots(i i) \end{array}$ From equation(i), write y in terms of $x$ to obtain $y=2 x+3$ On...
Find the roots of the given equation:
We write, $x=4 x-3 x$ as $2 x^{2} \times(-6)=-12 x^{2}=4 x \times(-3 x)$ $\begin{array}{l} \therefore 2 x^{2}+x-6=0 \\ \Rightarrow 2 x^{2}+4 x-3 x-6=0 \\ \Rightarrow 2 x(x+2)-3(x+2)=0 \\...