Excercise 2A

### 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. => $a\left(\frac{2}{3}\right)^{2}+7\left(\frac{2}{3}\right)+b=0$...

### 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 $\alpha \beta=12$ $\therefore f(x)=x^{2}-(\alpha+\beta) x+\alpha \beta$...

$f(x)=5 x^{2}-4-8 x$ $=5 x^{2}-8 x-4$ $=5 x^{2}-(10 x-2 x)-4$ $=5 x^{2}-10 x+2 x-4$ $=5 x(x-2)+2(x-2)$ $=(5 x+2)(x-2)$ $\therefore \mathrm{f}(\mathrm{x})=0 \Rightarrow(5... read more ### Find the zeroes of the quadratic polynomial and verify the relation between its zeroes and coefficients.$f(x)=4 x^{2}-4 x-3=4 x^{2}-(6 x-2 x)-3=4 x^{2}-6 x+2 x-3=2 x(2 x-3)+1(2 x-3)=(2 x+1)(2 x-3)\therefore \mathrm{f}(\mathrm{x})=0 \Rightarrow(2 \mathrm{x}+1)(2 \mathrm{x}-3)=0$... read more ### Find the zeroes of the quadratic polynomial and verify the relation between its zeroes and coefficients.$f(x)=x^{2}+3 x-10=x^{2}+5 x-2 x-10=x(x+5)-2(x+5)=(x-2)(x+5)\therefore \mathrm{f}(\mathrm{x})=0 \Rightarrow(\mathrm{x}-2)(\mathrm{x}+5)=0\Rightarrow x-2=0$or$x+5=0\Rightarrow...

$x^{2}-2 x-8=0$ $\Rightarrow \mathrm{x}^{2}-4 \mathrm{x}+2 \mathrm{x}-8=0$ $\Rightarrow x(x-4)+2(x-4)=0$ $\Rightarrow(x-4)(x+2)=0$ $\Rightarrow(x-4)=0$ or $(x+2)=0$ $\Rightarrow x=4$ or $x=-2$ Sum...
$x^{2}+7 x+12=0$ $\Rightarrow x^{2}+4 x+3 x+12=0$ $\Rightarrow x(x+4)+3(x+4)=0$ $\Rightarrow(x+4)(x+3)=0$ $\Rightarrow(x+4)=0$ or $(x+3)=0$ $\Rightarrow \mathrm{x}=-4$ or $\mathrm{x}=-3$ Sum of...