$\begin{array}{l} 10 x-\frac{1}{x}=3 \\ \Rightarrow 10 x^{2}-1=3 x \end{array}$ [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...

### 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, $7 x=5 x+2 x$ 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...

### 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) $-3$ Since $\alpha$ 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) $\frac{-2}{3}$ Assuming $\alpha$ and $\beta$ be the zeroes of $\mathrm{kx}^{2}+2 \mathrm{x}+3 \mathrm{k}$. Then $\alpha+\beta=\frac{-2}{k}$ and $\alpha \beta=3$...

### 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...

### If one zero of be the reciprocal of the other, then ? (a) 3 (b) (c) (d)

The correct option is option (a) $\mathrm{k}=3$ Let $\alpha$ 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) $a=-2, b=-6$ Since, $-2$ 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) Also,...

### Find the roots of the given equation:

We write, $-2 \sqrt{2} x=-3 \sqrt{2} x+\sqrt{2} x$ as $\sqrt{3} x^{2} \times(-2 \sqrt{3})=-6 x^{2}=(-3 \sqrt{2} x) \times(\sqrt{2} x)$ $\therefore \sqrt{3} x^{2}-2 \sqrt{2} x-2 \sqrt{3}=0$...

### 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 $(k-1) x^{2}+k x+1$, we have: $(k-1) \times(-4)^{2}+k \times(-4)+1=0$ $\Rightarrow 16 \mathrm{k}-16-4 \mathrm{k}+1=0$...

### 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) $\frac{-9}{2}$ Since, $\alpha$ 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)...

### 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 \\...

### If and are the zeros of , then the value of is (a) 5 (b) (c) 8 (d)

The correct option is option (b) $-5$ Since, $\alpha$ and $\beta$ be the zeroes of $\mathrm{x}^{2}+5 \mathrm{x}+8$. If $\alpha+\beta$ is the sum of the roots and $\alpha \beta$ is the product, then...

### 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 Let $\alpha$ 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) $x^{2}-\frac{1}{10} x-\frac{3}{10}$ the zeroes are $\frac{3}{5}$ and $\frac{-1}{2}$ Let $\alpha=\frac{3}{5}$ and $\beta=\frac{-1}{2}$ sum of the zeroes,...

### 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...

### A quadratic polynomial whose zeroes are 5 and -3, is (a) (b) (c) (d) none of these

The correct option is option (c) $x^{2}-2 x-15$ Since, the zeroes are 5 and $-3$. Let $\alpha=5$ and $\beta=-3$ sum of the zeroes, $\alpha+\beta=5+(-3)=2$ Also, product of the zeroes, $\alpha...

### 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) $x^{2}-3 x-10$ Since, sum of zeroes, $\alpha+\beta=3$ Also, product of zeroes, $\alpha \beta=-10$ $\therefore$ Required polynomial...

### The zeros of the polynomial are (a) (a) (c) (d) none of these

The correct option is option (a) $\frac{2}{3}, \frac{-1}{7}$ Let $f(x)=7 x^{2}-\frac{11}{3} x-\frac{2}{3}=0$ $\Rightarrow 21 \mathrm{x}^{2}-11 \mathrm{x}-2=0$ $\Rightarrow 21 \mathrm{x}^{2}-14...

### 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...

### The zeros of the polynomial are (a) (b) (c) (d) none of these

The correct option is option (b) $\frac{-3}{2}, \frac{4}{3}$ Let $f(x)=x^{2}+\frac{1}{6} x-2=0$ $\Rightarrow 6 \mathrm{x}^{2}+\mathrm{x}-12=0$ $\Rightarrow 6 x^{2}+9 x-8 x-12=0$ $\Rightarrow 3 x(2...

### The zeroes of the polynomial are (a) (b) (c) (d) none of these

The correct option is option (c) $\frac{3}{\sqrt{2}}, \frac{\sqrt{2}}{4}$ Let $f(x)=4 x^{2}+5 \sqrt{2} x-3=0$ $\Rightarrow 4 x^{2}+6 \sqrt{2} x-\sqrt{2} x-3=0$ $\Rightarrow 2 \sqrt{2}...

### 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...

### The zeroes of the polynomial are (a) (b) (c) (d)

The correct option is option (b) $3 \sqrt{2},-2 \sqrt{2}$ Let $f(x)=x^{2}-\sqrt{2} x-12=0$ $\Rightarrow x^{2}-3 \sqrt{2} x+2 \sqrt{2} x-12=0$ $\Rightarrow x(x-3 \sqrt{2})+2 \sqrt{2}(x-3 \sqrt{2})=0$...

### 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 $\mathrm{x}=-1$

### 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...

### 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 $a_{n} \neq 0$.

### 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 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 \\...

### 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 \\...

### Find the roots of the given equation:

$\begin{array}{l} 3 x^{2}-243=0 \\ \Rightarrow 3\left(x^{2}-81\right)=0 \\ \Rightarrow(x)^{2}-(9)^{2}=0 \\ \Rightarrow(x+9)(x-9)=0 \\ \Rightarrow x+9=0 \text { or } x-9=0 \\ \Rightarrow x=-9 \text {...

### Find the zeroes of the quadratic polynomial .

To find the zeroes of the quadratic polynomial we will equate $\mathrm{f}(\mathrm{x})$ to 0 Hence, the zeroes of the quadratic polynomial $f(x)=6 x^{2}-3$ are...

### Solve for x and y:

3x – 5y – 19 = 0,

-7x + 3y + 1 = 0

Solution: The given system of equation is: $\begin{array}{c} 3 \mathrm{x}-5 \mathrm{y}-19=0 \\ -7 \mathrm{x}+3 \mathrm{y}+1=0 \end{array}$ When multiplying equation(i) by 3 and equation(ii) by 5 ,...

### Find the roots of the given equation:

$\begin{array}{l} (2 x-3)(3 x+1)=0 \\ \Rightarrow 2 x-3=0 \text { or } 3 x+1=0 \\ \Rightarrow 2 x=3 \text { or } 3 x=-1 \\ \Rightarrow x=\frac{3}{2} \text { or } x=-\frac{1}{3} \end{array}$ As a...

### Solve for x and y:

2x – 3y = 13,

7x – 2y = 20

Solution: The given system of eq. is: $\begin{array}{cc} 2 \mathrm{x}-3 \mathrm{y}=13 & \ldots \ldots \text { (i) } \\ 7 \mathrm{x}-2 \mathrm{y}=20 & \ldots \ldots \text { (ii) }...

### Find the sum of the zeros and the product of zeros of a quadratic polynomial, are and respectively. Write the polynomial.

We can find the quadratic polynomial if we know the sum of the roots and product of the roots by using the formula $\mathrm{x}^{2}-($ sum of the zeroes $) \mathrm{x}+$ product of zeroes $\Rightarrow...

### State Division Algorithm for Polynomials.

"If $\mathrm{f}(\mathrm{x})$ and $\mathrm{g}(\mathrm{x})$ are two polynomials such that degree of $\mathrm{f}(\mathrm{x})$ is greater than degree of $\mathrm{g}(\mathrm{x})$ where...

### If and be the zeroes of the polynomial write the value of .

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 {...

### Solve for x and y:

2x + 3y = 0,

3x + 4y = 5

Solution: The given system of equation is: $\begin{array}{l} 2 x+3 y=0 & \ldots \ldots \text { (i) } \\ 3 x+4 y=5 & \ldots \text {..(ii) } \end{array}$ When multiplying equation(i) by 4 and...

### If is divisible by , write the value of a and .

Equating $\mathrm{x}^{2}-\mathrm{x}$ to 0 to find the zeroes, we will get x(x-1)=0 x(x-1)=0 ⇒x=0 or x-1=0 \Rightarrow \mathrm{x}=0 \text { or } \mathrm{x}-1=0...

### If and are zeros of the polynomial write the value of a.

using the relationship between the zeroes of the quadratic polynomial.$$ \begin{aligned} &\text { Sum of zeroes }=\frac{-\left(\text { coefficient of } x^{2}\right)}{\text { coefficient of } x^{3}}...

### If is a factor of , then find the value of

$(x+a)$ is a factor of $2 x^{2}+2 a x+5 x+10$ $x+a=0$ $\Rightarrow \mathrm{X}=-\mathrm{a}$ Since, $(x+a)$ is a factor of $2 x^{2}+2 a x+5 x+10$ Hence, It will satisfy the above polynomial...

### Solve for x and y:

x-y=3

Solution: The given system of equations is $x-y=3 \dots \dots(i)$ $\frac{x}{3}+\frac{y}{2}=6 \dots \dots(ii)$ From equation(i), write y in terms of $x$ to obtain $y=x-3$ When substituting...

### If the product of the zero of the polynomial is 3 . Find the value of .

using the relationship between the zeroes of he quadratic polynomial. Product of zeroes $=\frac{\text { constant term }}{\text { coefficient of } x^{2}}$ $\Rightarrow 3=\frac{k}{1}$ $\Rightarrow...

### If the sum of the zeros of the quadratic polynomial is 1 write the value of .

using the relationship between the zeroes of the quadratic polynomial. Sum of zeroes $=\frac{-(\text { coef ficient of } x)}{\text { coefficient of } x^{2}}$ $\Rightarrow 1=\frac{-(-3)}{k}$...

### Write the zeros of the polynomial .

$f(x)=x^{2}-x-6$ $=x^{2}-3 x+2 x-6$ $=x(x-3)+2(x-3)$ $=(x-3)(x+2)$ $f(x)=0 \Rightarrow(x-3)(x+2)=0$ $$ \begin{aligned} &\Rightarrow(x-3)=0 \text { or }(x+2)=0 \\ &\Rightarrow x=3 \text { or } x=-2...

### If is a zero of the polynomial then find the value of .

$x=-2$ is one zero of the polynomial $3 x^{2}+4 x+2 k$ Therefore, it will satisfy the above polynomial. Now, we have $3(-2)^{2}+4(-2) 1+2 k=0$ $\Rightarrow 12-8+2 k=0$ $\Rightarrow...

### If 1 is a zero of the quadratic polynomial is 1 , then find the value of a.

$x=1$ is one zero of the polynomial $a x^{2}-3(a-1) x-1$ Therefore, it will satisfy the above polynomial. Now, we have $a(1)^{2}-(a-1) 1-1=0$ $\Rightarrow a-3 a+3-1=0$ $\Rightarrow-2 \mathrm{a}=-2$...

### Solve for x and y:

x + y = 3,

4x – 3y = 26

Solution: The given system of equation is: $\mathrm{x}+\mathrm{y}=3 \ldots \ldots \text { (i) }$ $4 x-3 y=26 \ldots \ldots \text { (ii) }$ When multiplying equation(i) by 3 , we obtain: $3 x+3 y=9...

### If is a zero of the polynomial is , then find the value of .

$x=-4$ is one zero of the polynomial $x^{2}-x-(2 k+2)$ Therefore, it will satisfy the above polynomial. Now, we have $(-4)^{2}-(-4)-(2 k+2)=0$ $\Rightarrow 16+4-2 \mathrm{k}-2=0$ $\Rightarrow 2...

### If 3 is a zero of the polynomial , find the value of .

$x=3$ is one zero of the polynomial $2 x^{2}+x+k$ Therefore, it will satisfy the above polynomial. Now, we have $2(3)^{2}+3+k=0$ $\Rightarrow 21+\mathrm{k}=0$ $\Rightarrow...

### If one zero of the quadratic polynomial is 2 , then find the value of .

$x=2$ is one zero of the quadratic polynomial $k x^{2}+3 x+k$ Therefore, it will satisfy the above polynomial. $k(2)^{2}+3(2)+k=0$ $\Rightarrow 4 \mathrm{k}+6+\mathrm{k}=0$ $\Rightarrow 5...

### Find the value of for which is a root of the equation .

It is given that $(x=1)$ is a root of $\left(x^{2}+k x+3=0\right)$. As a result, $(x=1)$ must satisfy the equation. $\begin{array}{l} \Rightarrow(1)^{2}+k \times 1+3=0 \\ \Rightarrow k+4=0 \\...

### Find are the zeros of polynomial and then write the polynomial.

If the zeroes of the quadratic polynomial are $\alpha$ and $\beta$ then the quadratic polynomial can be found as $\mathrm{x}^{2}-(\alpha+\beta) \mathrm{x}+\alpha \beta$ $\ldots \ldots(1)$...

### Draw the graphs for the following equations on the same graph paper:

2x + y = 2

2x + y = 6 Find the co-ordinates of the vertices of the trapezium formed by these lines. Also, find the area of the trapezium so formed.

Solution: From the first eq., write $y$ in terms of $x$ $y=2-2 x\dots \dots(i)$ Substituting different values of $\mathrm{x}$ in equation(i) to get different values of $\mathrm{y}$ For $x=0,...

### Find the zeroes of the polynomial

$f(x)=x^{2}-3 x-m(m+3)$ adding and subtracting $\mathrm{mx}$, $f(x)=x^{2}-m x-3 x+m x-m(m+3)$ $=x[x-(m+3)]+m[x-(m+3)]$ $=[x-(m+3)](x+m)$ $f(x)=0 \Rightarrow[x-(m+3)](x+m)=0$...

### Which of the following are the roots of

On subtracting $x=\left(-\frac{1}{2}\right)$ in the given equation, we get $\begin{array}{l} \text { L.H.S. }=3 x^{2}+2 x-1 \\ =3 \times\left(-\frac{1}{2}\right)^{2}+2...

### Find the zeroes of the polynomial

$f(x)=x^{2}+x-p(p+1)$ adding and subtracting $\mathrm{px}$, we get $f(x)=x^{2}+p x+x-p x-p(p+1)$ $=x^{2}+(p+1) x-p x-p(p+1)$ $=x[x+(p+1)]-p[x+(p+1)]$ $=[x+(p+1)](x-p)$ $f(x)=0$...

### If one zero of the polynomial Is , write the other zero.

Let the other zeroes of $x^{2}-4 x+1$ be a (using the relationship between the zeroes of the quadratic polynomial) sum of zeroes $=\frac{-(\text { coef ficient of } x)}{\text { coef ficient of }...

### Show graphically that the system of equations 2x + y = 6, 6x + 3y = 20 is inconsistent.

Solution: From the first equation, write y in terms of $x$ $y=6-2 x\dots \dots(i)$ Substituting different values of $x$ in equation(i) to get different values of $y$ For $x=0, y=6-0=6$ For $x=2,...

### Find all the zeroes of polynomial , it being given that two of its zeroes are and .

$f(x)=2 x^{4}-11 x^{3}+7 x^{2}+13 x-7$. Since $(3+\sqrt{2})$ and $(3-\sqrt{2})$ are the zeroes of $f(x)$ it follows that each one of $(x+3+\sqrt{2})$ and $(x+3-\sqrt{2})$ is a factor of $f(x)$...

### Which of the following are quadratic equation in ?

(i)

(ii)

(i) $\begin{array}{l} (x+2)^{3}=x^{3}-8 \\ \Rightarrow x^{3}+6 x^{2}+12 x+8=x^{3}-8 \\ \Rightarrow 6 x^{2}+12 x+16=0 \end{array}$ This is of the form $a x^{2}+b x+c=0$ Hence, the given equation is a...

### Obtain all other zeroes of if two of its zeroes are and .

The given polynomial is $f(x)=x^{4}+4 x^{3}-2 x^{2}-20 x-15$. Since $(x-\sqrt{5})$ and $(x+\sqrt{5})$ are the zeroes of $f(x)$ it follows that each one of $(x-\sqrt{5})$ and $(x$ $+\sqrt{5})$ is a...

### Show graphically that the system of equations 2x + 3y = 4, 4x + 6y = 12 is inconsistent.

Solution: Draw a horizontal line on a graph paper X'OX and a vertical line YOY' as the x-axis and y-axis, respectively. Graph of $2 x+3 y=4$ $\begin{array}{l} 2 x+3 y=4 \\ \Rightarrow 3 y=(-2...

### Find all the zeroes of , it is being given that two of its zeroes are and .

The given polynomial is $f(x)=2 x^{4}-3 x^{3}-5 x^{2}+9 x-3$ Since $\sqrt{3}$ and $-\sqrt{3}$ are the zeroes of $f(x)$, it follows that each one of $(x-\sqrt{3})$ and $(x+\sqrt{3})$ is a factor of...

### Which of the following are quadratic equation in ?

(i)

(ii)

(i) Clearly, $\left(\sqrt{2} x^{2}+7 x+5 \sqrt{2}\right)$ is a quadratic polynomial. $\therefore \sqrt{2} x^{2}+7 x+5 \sqrt{2}=0$ is a quadratic equation. (ii) Clearly, $\left(\frac{1}{3}...

### Show graphically that the system of equations x – 2y = 6, 3x – 6y = 0 is inconsistent.

Solution: From the first eq., write y in terms of $x$ $\mathrm{y}=\frac{x-6}{2}\dots \dots(i)$ Substituting different values of $x$ in equation(i) to get different values of y For $x=-2,...

### Which of the following are quadratic equation in ?

(i)

(ii)

(i) $\left(x^{2}-x+3\right)$ is a quadratic polynomial $\therefore x^{2}-x+3=0$ is a quadratic equation. (ii) Clearly, $\left(2 x^{2}+\frac{5}{2} x-\sqrt{3}\right)$ is a quadratic polynomial....

### Show graphically that the system of equations x – 2y = 5, 3x – 6y = 15 has infinitely many solutions.

Solution: From the first eq., write y in terms of $x$ $y=\frac{x-5}{2}\dots \dots(i)$ Substituting different values of $x$ in eq.(i) to get different values of y For $x=-5, y=\frac{-5-5}{2}=-5$ For...

### Show graphically that the system of equations 2x + y = 6, 6x + 3y = 18 has infinitely many solutions.

Solution: Draw a horizontal line on a graph paper $\mathrm{X}^{\prime} \mathrm{OX}$ and a vertical line YOY' representing the $\mathrm{x}-$ axis and $y$-axis, respectively. Graph of $2...

### Show graphically that the system of equations 3x – y = 5, 6x – 2y = 10 has infinitely many solutions.

Solution: Draw a horizontal line on a graph paper X'OX and a vertical line YOY' representing the xaxis and y-axis, respectively. Graph of $3 x-y=5$ $3 x-y=5$ $\Rightarrow y=3 x-5\dots \dots(i)$...

### Show graphically that the system of equations 2x + 3y = 6, 4x + 6y = 12 has infinitely many solutions.

Solution: From the first eq., write y in terms of $x$ $y=\frac{6-2 x}{3}\dots \dots(i)$ Substituting different values of $x$ in equation(i) to get different values of y For $x=-3, y=\frac{6+6}{3}=4$...

### Find all the zeroes of , if it is given that two of its zeroes are and

Let f(x)=x4+x3-23x2-3x+60 \text { Let } f(x)=x^{4}+x^{3}-23 x^{2}-3 x+60 Since $\sqrt{3}$ and $-\sqrt{3}$ are the zeroes of $f(x)$, it follows that each one of $(x-\sqrt{3})$ and...

### If 2 and are two zeroes of the polynomial , find all the zeroes of the given polynomial.

Let $f(x)=x^{4}+x^{3}-34 x^{2}-4 x+120$ Since 2 and $-2$ are the zeroes of $f(x)$, it follows that each one of $(x-2)$ and $(x+2)$ is a factor of $f(x)$ Consequently,...

### If 3 and are two zeroes of the polynomial , find all the zeroes of the given polynomial.

Since 3 and $-3$ are the zeroes of $f(x)$, it follows that each one of $(x+3)$ and $(x-3)$ is a factor of $f(x)$ Consequently, $(x-3)(x+3)=\left(x^{2}-9\right)$ is a factor of $f(x)$. On dividing...

### If 1 and are two zeroes of the polynomial , find its third zero.

Let $f(x)=x^{3}-4 x^{2}-7 x+10$ Since 1 and $-2$ are the zeroes of $f(x)$, it follows that each one of $(x-1)$ and $(x+2)$ is a factor of $f(x)$ Consequently, $(x-1)(x+2)=\left(x^{2}+x-2\right)$ is...

### It is given that is one of the zeroes of the polynomial . Find all the zeroes of the given polynomial.

Let $f(x)=x^{3}+2 x^{2}-11 x-12$ Since $-1$ is a zero of $f(x),(x+1)$ is a factor of $f(x)$. On dividing $\mathrm{f}(\mathrm{x})$ by $(\mathrm{x}+1)$, we get $$ \begin{aligned} &f(x)=x^{3}+2...

### On dividing is divided by a polynomial , the quotient and remainder are and respectively. Find

using division rule, Dividend $=$ Quotient $\times$ Divisor $+$ Remainder $\therefore 3 x^{3}+x^{2}+2 x+5=(3 x-5) g(x)+9 x+10$ $\Rightarrow 3 x^{3}+x^{2}+2 x+5-9 x-10=(3 x-5) g(x)$ $\Rightarrow 3...

### By actual division, show that is a factor of .

Let $f(x)=2 x^{4}+3 x^{3}-2 x^{2}-9 x-12$ and $g(x)$ as $x^{2}-3$

### Solve graphically the system of equations

2x – 3y = 12

x + 3y = 6. Find the coordinates of the vertices of the triangle formed by these two lines and the y-axis.

Solution: Draw a horizontal line on a graph paper $X^{\prime} O X$ and a vertical line YOY' as the $x$-axis and $y$-axis, respectively. $\text { Graph of } 4 x-3 y+4=0$ $\begin{array}{l} 4 x-3 y+4=0...

### Solve graphically the system of equations

5x – y = 7

x – y + 1 = 0. Find the coordinates of the vertices of the triangle formed by these two lines and the y-axis.

Solution: From the first eq., write y in terms of $x$ $y=\frac{2 x-12}{3}\dots \dots (i)$ Substituting different values of $\mathrm{x}$ in eq.(i) to get different values of $\mathrm{y}$ For $x=0,...

### Solve graphically the system of equations

2x – 5y + 4 = 0

2x + y – 8 = 0. Find the coordinates of the vertices of the triangle formed by these two lines and the y-axis.

Solution: Draw a horizontal line on a graph paper $X^{\prime} O X$ and a vertical line YOY' as the $x$-axis and $y$-axis, respectively. $\begin{array}{l} \quad \text { Graph of } 2 x-5 y+4=0 \\ 2...

### Solve graphically the system of equations

x – y – 5 = 0

3x + 5y – 15 = 0. Find the coordinates of the vertices of the triangle formed by these two lines and the y-axis.

Solution: Draw a horizontal line on a graph paper $X^{\prime} O X$ and a vertical line YOY' as the $x$-axis and $y$-axis, respectively. $\text { Graph of } 2 x-y=1$ $\begin{array}{l} 2 x-y=1 \\...

### Solve graphically the system of equations

4x – y – 4 = 0

3x + 2y – 14 = 0. Find the coordinates of the vertices of the triangle formed by these two lines and the y-axis.

Solution: Draw a horizontal line on a graph paper $X^{\prime} O X$ and a vertical line $Y O Y^{\prime}$ as the $x$-axis and $y$-axis, respectively. Graph of $4 x-y=4$ $\begin{array}{l} 4 x-y=4 \\...

### Solve graphically the system of equations

2x – 3y + 6 = 0

2x + 3y – 18 = 0. Find the coordinates of the vertices of the triangle formed by these two lines and the y-axis.

Solution: Draw a horizontal line on a gfraph paper $X^{\prime} O X$ and a vertical line $Y O Y^{\prime}$ as the $x$-axis and $y$-axis, respectively. Graph of $2 x-3 y-17=0$ $\begin{array}{l} 2 x-3...

### Solve the following system of equations graphically: 4x-5y+16=0, 2x+y-6=0. Determine the vertices of the triangle formed by these lines and the x-axis.

Solution: On a graph paper, draw a horizontal line $X^{\prime} O X$ and a vertical line YOY' as the $x$-axis and $y$-axis, respectively. Graph of $4 x-5 y+16=0$ $\begin{array}{l} 4 x-5 y+16=0 \\...

### Solve the following system of linear equations graphically

x-y+1=0,

3 x+2 y-12=0 Calculate the area bounded by these lines and the -axis.f

Solution: Draw a horizontal line on a graph paper $\mathrm{X}^{\prime} \mathrm{OX}$ and a vertical line YOY' as the $\mathrm{x}$-axis and $\mathrm{y}$-axis, respectively. Graph of...

### Solve the following system of linear equations graphically

4 x-3 y+4=0,

4 x+3 y-20=0 Find the area of the region bounded by these lines and the x-axis.

Solution: Draw a horizontal line on a graph paper X'OX and a vertical line YOY' as the $\mathrm{x}$-axis and $\mathrm{y}$-axis, respectively. Graph of $4 x-3 y+4=0$ $\begin{array}{l} 4 x-3 y+4=0 \\...

### Solve graphically the system of equations

2x-3y+4=0

x+2y-5=0 Find the coordinates of the vertices of the triangle formed by these two lines and the -axis.

Solution: From the first eq., write y in terms of $x$ $\mathrm{y}=\frac{2 x+4}{3}\dots \dots(i)$ Substituting different values of $x$ in eq.(i) to get different values of $y$ For $x=-2,...

### Solve graphically the system of equations

x-y-3=0

2x-3y-4=0 Find the coordinates of the vertices of the triangle formed by these two lines and the -axis.

Solution: From the first eq., write y in terms of $\mathrm{x}$ $y=x+3\dots \dots(i)$ Substituting different values of $x$ in eq(i) to get different values of $y$ For $x=-3, y=-3+3=0$ For $x=-1,...

### If is divided by .

$f(x)$ as $x^{4}+0 x^{3}+0 x^{2}-5 x+6$ and $g(x) a s-x^{2}+2$ Quotient $q(x)=-x^{2}-2$ Remainder $\mathrm{r}(\mathrm{x})=-5 \mathrm{x}+10$

### An analysis of the weekly wages paid to workers in two firms A and B, belonging to the same industry gives the following results:

(i) Which firm A or B pays out the larger amount as weekly wages? (ii) Which firm A or B has greater variability in individual wages? Solution:

### If f(x) =

is divided by g(x)=

Quotient $q(x)=x^{2}+x-3$ Remainder $r(x)=8$

### If is divided by

Quotient $q(x)=x-\overline{3}$ Remainder $r(x)=7 x-9$ 7. If $f(x)=x^{4}-3 x^{2}+4 x+5$ is divided by $g(x)=x^{2}-x+1$

### Find a cubic polynomial with the sum of its zeroes, sum of the products of its zeroes taken two at a time and the product of its zeroes as and respectively.

sum of the product of the zeroes taken two at a time and the product of the zeroes of a cubic polynomial then the cubic polynomial can be found as $x^{3}-($ sum of the zeroes $) x^{2}+($ sum of the...

### Find a cubic polynomial whose zeroes are and .

If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as $x^{3}-(a+b+c) x^{2}+(a b+b c+c a) x-a b c$ Let $a=\frac{1}{2}, b=1$ and $c=-3$ Substituting the values...

### Find a cubic polynomial whose zeroes are and

If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as $x^{3}-(a+b+c) x^{2}+(a b+b c+c a) x-a b c$ Let $a=2, b=-3$ and $c=4$ Substituting the values in 1 , we...

### Verify that and are the zeroes of the cubic polynomial and verify the relation between its zeroes and coefficients.

p(x)=3x3-10x2-27x+10 p(x)=\left(3 x^{3}-10 x^{2}-27 x+10\right) p(5)=3×53-10×52-27×5+10=(375-250-135+10)=0 p(5)=\left(3 \times 5^{3}-10 \times 5^{2}-27 \times...

### Verify that are the zeros of the cubic polynomial and verify the relation between it zeros and coefficients.

The given polynomial is $p(x)=\left(x^{3}-2 x^{2}-5 x+6\right)$ $$ \begin{aligned} &\therefore \mathrm{p}(3)=\left(3^{3}-2 \times 3^{2}-5 \times 3+6\right)=(27-18-15+6)=0 \\...

### One zero of the polynomial is . Find the other zeros of the polynomial.

$x=\frac{2}{3}$ is one of the zero of $3 x^{3}+16 x^{2}+15 x-18$ Now, we have $\mathrm{x}=\frac{2}{3}$ $\Rightarrow \mathrm{x}-\frac{2}{3}=0$ Now, we divide $3 x^{3}+16 x^{2}+15 x-18$ by...

### Solve the system of equations graphically:

x+2y+2=0

3x+2y-2=0

Solution: Draw a horizontal line on a graph paper X'OX and a vertical line YOY' as the $\mathrm{x}$-axis and $\mathrm{y}$-axis, respectively. $\begin{array}{l} x+2 y+2=0 \\ \Rightarrow 2 y=(-2-x) \\...

### Solve the system of equations graphically:

2x+3y=4

3x-y=-5

Solution: Draw a horizontal line oon a graph paper X'OX and a vertical line YOY' as the $x$-axis and $y$-axis, respectively. Graph of $2 x+3 y=4$ $2 x+3 y=4$ $\Rightarrow 3 y=(4-2 x)$ $\therefore...

### If is a factor of the polynomial , find the value of .

$(x+a)$ is a factor of $2 x^{2}+2 a x+5 x+10$ $x+a=0$ $\Rightarrow \mathrm{x}=-\mathrm{a}$ Since, it satisfies the above polynomial. => $2(-a)^{2}+2 a(-a)+5(-a)+10=0$ $\Rightarrow 2 a^{2}-2...

### If and are the roots of the quadratic equation then find the value of a and .

$a x^{2}+7 x+b=0$ Since, $x=\frac{2}{3}$ is the root of the above quadratic equation Hence, it will satisfy the above equation. => $a\left(\frac{2}{3}\right)^{2}+7\left(\frac{2}{3}\right)+b=0$...

### Solve the system of equations graphically:

2x-3y+13=0

3x-2y+12=0

Solution: From the first eq., write y in terms of $x$ $y=\frac{2 x+13}{3}\dots \dots (i)$ Substituting different values of $x$ in eq.(i) to get different values of $y$ For $x=-5,...

### Find the quadratic polynomial, sum of whose zeroes is and their product is .

Quadratic equation can be found if we know the sum of the roots and product of the roots by using the formula: $\mathrm{x}^{2}-($ Sum of the roots) $\mathrm{x}+$ Product of roots $=0$ $\Rightarrow...

### Solve the system of equations graphically:

2x+3y+5=0

3x-2y-12=0

Solution: From the first eq., write y in terms of $x$ $y=-\left(\frac{5+2 x}{3}\right)\dots \dots (i)$ Substitute different values of $\mathrm{x}$ in eq.(i) to get different values of $\mathrm{y}$...

### Find the quadratic polynomial, sum of whose zeroes is and their product is 1 . Hence, find the zeroes of the polynomial.

Let $\alpha$ and $\beta$ be the zeroes of the required polynomial $\mathrm{f}(\mathrm{x})$. =>$(\alpha+\beta)=\frac{5}{2}$ and $\alpha \beta=1$ $\therefore...

### Solve the system of equations graphically:

3x+y+1=0

2x-3y+8=0

Solution: Draw a horizontal line on a graph paper X'OX and a vertical line YOY' as the $\mathrm{x}$-axis and y-axis, respectively. Graph of $3 \mathbf{x}+\mathbf{y}+\mathbf{1}=\mathbf{0}$...

### Find the quadratic polynomial, sum of whose zeroes is 0 and their product is . Hence, find the zeroes of the polynomial.

Let $\alpha$ and $\beta$ be the zeroes of the required polynomial $f(x)$. Then $(\alpha+\beta)=0$ and $\alpha \beta=-1$ $\therefore f(x)=x^{2}-(\alpha+\beta) \mathrm{x}+\alpha \beta$ $\Rightarrow...

### Solve the system of equations graphically:

3x+2y=12

5x-2y=4

Solution: The given eq. are: $3 x+2 y=12 \dots\dots (i)$ $5 x-2 y=4 \dots \dots(ii)$ From eq.(i), write y in terms of $x$ $\mathrm{y}=\frac{12-3 x}{2}\dots \dots(iii)$ Substitute different values of...

### Solve the system of equations graphically:

2x-5y+4=0

2x+y-8=0

Solution: Draw a horizontal line on a graph paper X'OX and a vertical line YOY' as the $\mathrm{x}$-axis and y-axis, respectively. $\begin{array}{l} 2 x-5 y+4=0 \\ \Rightarrow 5 y=(2 x+4) \\...

### Solve the system of equations graphically:

2x+3y=8

x-2y+3=0

Solution: Draw a horizontal line on a graph paper X'OX and a vertical line YOY' as the $\mathrm{x}$-axis and $\mathrm{y}$-axis, respectively. $\begin{array}{l} 2 x+3 y=8 \\ \Rightarrow 3 y=(8-2 x)...

### Find the quadratic polynomial, sum of whose zeroes is 8 and their product is Hence, find the zeroes of the polynomial.

Let $\alpha$ and $\beta$ be the zeroes of the required polynomial $\mathrm{f}(\mathrm{x})$. Then $(\alpha+\beta)=8$ and $\alpha \beta=12$ $\therefore f(x)=x^{2}-(\alpha+\beta) x+\alpha \beta$...

### Find the quadratic polynomial whose zeroes are and . Verify the relation between the coefficients and the zeroes of the polynomial.

Let $\alpha=\frac{2}{3}$ and $\beta=\frac{-1}{4}$. Sum of the zeroes $=(\alpha+\beta)=\frac{2}{3}+\left(\frac{-1}{4}\right)=\frac{8-3}{12}=\frac{5}{12}$ Product of the zeroes, $\alpha...

### Solve the system of equations graphically:

3 x+2 y=4

2 x-3 y=7

Solution: Draw a horizontal line on a graph paper X'OX and a vertical line YOY' representing the xaxis and $y$-axis, respectively. Graph of $3 x+2 y=4$ $3 x+2 y=4$ $\Rightarrow 2 y=(4-3 x)$...

### Solve the system of equations graphically:

2x + 3y = 2,

x – 2y = 8

Solution: Draw a horizontal line on a graph paper$\mathrm{X}^{\prime} \mathrm{OX}$ and a vertical line YOY' representing the $\mathrm{x}-$ axis and $\mathrm{y}$-axis, respectively. $\begin{array}{l}...

### Find the zeroes of the quadratic polynomial and verify the relation between the zeroes and the coefficients.

$$ \begin{aligned} &3 x^{2}-x-4=0 \\ &\Rightarrow 3 x^{2}-4 x+3 x-4=0 \\ &\Rightarrow x(3 x-4)+1(3 x-4)=0 \\ &\Rightarrow(3 x-4)(x+1)=0 \\ &\Rightarrow(3 x-4) \text { or }(x+1)=0 \\ &\Rightarrow...

### Find the zeroes of the quadratic polynomial (5y and verify the relation between the zeroes and the coefficients.

f(u)=5u2+10u \mathrm{f}(\mathrm{u})=5 \mathrm{u}^{2}+10 \mathrm{u} It can be written as $5 \mathrm{u}(\mathrm{u}+2)$ ∴f(u)=0⇒5u=0 or u+2=0 \therefore \mathrm{f}(\mathrm{u})=0...

### Find the zeroes of the quadratic polynomial and verify the relation between the zeroes and the coefficients.

$$ \begin{aligned} &\mathrm{f}(\mathrm{x})=8 \mathrm{x}^{2}-4 \\ &\text { It can be written as } 8 \mathrm{x}^{2}+0 \mathrm{x}-4 \\ &=4\left\{(\sqrt{2} x)^{2}-(1)^{2}\right\} \\ &=4(\sqrt{2}...

### Find the zeroes of the quadratic polynomial and verify the relation between the zeroes and the coefficients.

$$ \begin{aligned} &4 x^{2}-4 x+1=0 \\ &\Rightarrow(2 x)^{2}-2(2 x)(1)+(1)^{2}=0 \end{aligned} $$ $$ \begin{aligned} &\Rightarrow(2 \mathrm{x}-1)^{2}=0 \quad\left[\because \mathrm{a}^{2}-2...

### Find the standard deviation for the following distribution:

Solution: By using the formula for standard deviation: $\begin{array}{l} \mathrm{SD}=\sqrt{\operatorname{Var}(\mathrm{X})} \\ \text { Mean }=\sum \frac{\mathrm{f}_{\mathrm{i}}...

$$ \begin{aligned} \mathrm{f}(\mathrm{x}) &=2 \mathrm{x}^{2}-11 \mathrm{x}+15 \\ &=2 \mathrm{x}^{2}-(6 \mathrm{x}+5 \mathrm{x})+15 \\ &=2 \mathrm{x}^{2}-6 \mathrm{x}-5 \mathrm{x}+15 \\ =& 2...

### Find the zeroes of the polynomial and verify the relation between its zeroes and coefficients.

$$ \begin{aligned} &2 \sqrt{3} x^{2}-5 x+\sqrt{3} \\ &\Rightarrow 2 \sqrt{3} x^{2}-2 x-3 x+\sqrt{3} \\ &\Rightarrow 2 x(\sqrt{3} x-1)-\sqrt{3}(\sqrt{3} x-1)=0 \\ &\Rightarrow(\sqrt{3} x-1) \text {...

### Find the zeros of the polynomial and verify the relation between its zeroes and coefficients.

$x^{2}+7 x+12=0$ $\Rightarrow x^{2}+4 x+3 x+12=0$ $\Rightarrow x(x+4)+3(x+4)=0$ $\Rightarrow(x+4)(x+3)=0$ $\Rightarrow(x+4)=0$ or $(x+3)=0$ $\Rightarrow \mathrm{x}=-4$ or $\mathrm{x}=-3$ Sum of...

### Show that every positive even integer is of the form (6m+1) or (6m+3) or (6m+5)where m is some integer.

Solution: Let's say that $n$ be any arbitrary positive odd integer. Upon dividing $n$ by 6, let $m$ be the quotient and $r$ be the remainder. Therefore, by Euclid's division lemma, we get $n=6 m+r$,...

### Show that every positive integer is either even or odd?

Solution: Let's suppose that there exist a smallest positive integer which is neither odd nor even, say $n .$ As $n$ is least positive integer which is neither even nor odd, $n-1$ must be either odd...

### Using Euclid’s algorithm, find the HCF of

(i) 405 and 2520

(ii) 504 and 1188

Solution: (i) When applying Euclid's algorithm, that is dividing $2520$ by $405$, we obtain, Quotient $=6$, Remainder $=90$ $\therefore 2520=405 \times 6+90$ Again upon applying Euclid's algorithm,...

### By what number should be 1365 be divided to get 31 as quotient and 32 as remainder?

Solution: It is given that, Dividend $=1365$, Quotient $=31$, Remainder Let the divisor be $x$ $\begin{array}{l}\text { Dividend }=\text { Divisor } \times \text { Quotient }+\text { Remainder } \\...

### A number when divided by 61 gives 27 as quotient and 32 as remainder. Find the number.

Solution: It is known to us that, Dividend $=$ Divisor $\times$ Quotient $+$ Remainder It is given that, Divisor $=61$, Quotient $=27$, Remainder $=32$ Let the Dividend be $x$ $\begin{aligned}...

### What do you mean by Euclid’s division algorithm?

Solution: It is stated by the Euclid's division algorithm that for any two positive integers $a$ and $b$, there exist unique integers $q$ and $r$, such that $a=b q+r$. where $0 \leq r \leq b$.

### Using prime factorization, find the HCF and LCM of (i)96, 404 (ii)144, 198 In each case verify that HCF × LCM = product of given numbers

Answers: (i) Using prime factorization, 96 = 25 × 3 404 = 22 × 101 HCF = 22 HCF = 4 LCM = 25 × 3 × 101 LCM = 9696 (ii) Using prime factorization, 144 = 24 × 32 198 = 2 × 32 × 11 HCF = 2 × 32 HCF =...

### Find the equations to the straight lines passing through the point (2, 3) and inclined at an angle of 450 to the lines 3x + y – 5 = 0.

The equation passes through (2, 3) and make an angle of 450with the line 3x + y – 5 = 0. Since, the equations of two lines passing through a point x1,y1 and making an angle α with the given line y =...

### Find the area between the curves and

Solution: The area required is represented by the shaded area OBAO as The point of intersection of the curves $y=x$ and $y=x^{2}$ is A(1,1) Draw AC perpendicular to x-axis $\therefore$ Area...

### The pressure of a 1:4 mixture of dihydrogen and dioxygen enclosed in a vessel is one atmosphere. What would be the partial pressure of dioxygen? (i) 0.8×105 atm (ii) 0.008 Nm–2 (iii) 8×104 Nm–2 (iv) 0.25 atm

The correct option is (iii) 8×104 Nm–2

### 4. If

, compute

Solution: Given that, \[sin\text{ }A\text{ }=\text{ }9/41~\ldots \ldots \ldots \ldots .\text{ }\left( 1 \right)\] Required to find: \[cos\text{ }A,\text{ }tan\text{ }A\] By definition, we...

### 11. Show that any positive odd integer is of the form 6q +1 or 6q + 3 or 6q + 5, where q is some integer.

Let ‘a’ be any positive integer. Then from Euclid’s division lemma, a = bq+r; where 0 < r < b Putting b=6 we get, ⇒ a = 6q + r, 0 < r < 6 For r = 0, we get\[a=6q=2\left( 3q...

**Solve each of the following systems of equations by the method of cross-multiplication:**

\[\mathbf{x}\text{ }+\text{ }\mathbf{2y}\text{ }+\text{ }\mathbf{1}\text{ }=\text{ }\mathbf{0}\] \[\mathbf{2x}\text{ }\text{ }\mathbf{3y}\text{ }\text{ }\mathbf{12}\text{ }=\text{...

**Find the distance between the following pairs of points:**

(i) (2, 3), (4, 1) (ii) (-5, 7), (-1, 3)(iii) (a, b), (- a, – b) We know that formula to find the distance (d) between two points $\left( {{x}_{1}},{{y}_{1}} \right)$and $\left( {{x}_{2}},{{y}_{2}}...