Question

Verify the quadratic polynomial, for the zeros $\alpha ,\beta$ given in each case

• A. $2,-1,x^2-x-2$
• B. $\sqrt{3},-\sqrt{3},x^2-4$
• C. $\frac{1}{4},-1,4x^2+3x-2$
• D. $\frac{1}{2},\frac{3}{2},4x^2-8x+4$

Answer 1

The correct option is A $2,-1,x^2-x-2$

(i) $2,-1$

$\Rightarrow (x-\alpha )(x-\beta )=0$

$\Rightarrow (x-2)(x-(-1\left)\right)=0$

$\Rightarrow (x-2)(x+1)=x^2-x-2$ is required quadratic polynomical

(ii) $\sqrt{3},-\sqrt{3}$

$\Rightarrow$$(x-\sqrt{3})(x+\sqrt{3})=0$

$\Rightarrow$$x^2-(\sqrt{3}{)}^2\to x^2-3=0$

$\Rightarrow$$x^2-3=0$ required quadratic polynomical

(iii) $\frac{1}{-4},-1$

$\Rightarrow$$(x-\frac{1}{4})(x-(-1\left)\right)=0$

$\Rightarrow$$x^2+x-\frac{1}{4}x-\frac{1}{4}=0$

$\Rightarrow$$x^2+\frac{3}{4}x-\frac{1}{4}=0$

$\Rightarrow$ $4x^2+3x-1=0$ is required quadratic equation

(iv) $\frac{1}{2},\frac{3}{2}$

$\Rightarrow (x-\frac{1}{2})(x-\frac{3}{2})$

$\Rightarrow$$x^2-(\frac{1}{2}+\frac{3}{2})x+\frac{3}{4}=0$

$=x^2-2x+\frac{3}{4}=0$

$\Rightarrow 4x^2-8x+3=0$ in required quadratic equation.