RS Aggarwal

### 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 $\alpha \beta=3$...

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

### 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...

### 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...

### 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 $\mathrm{x}^{2}-(\alpha+\beta) \mathrm{x}+\alpha \beta$ $\ldots \ldots(1)$...

### Find the zeroes of the polynomial

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

### 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 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...

### 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: 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 $3 \mathbf{x}+\mathbf{y}+\mathbf{1}=\mathbf{0}$...

### 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 $3 x+2 y=4$ $3 x+2 y=4$ $\Rightarrow 2 y=(4-3 x)$...