Question

Find the quadratic polynomial whose zeros are 2/3 and -1/4. Verify the relation between the coefficients and the zeros of the polynomial.

Answer 1

$\bold\red{\underline{\underline{Answer:}}}$

$\bold\green{\underline{\underline{Solution}}}$

$\bold{Zeroes \ of \ polynomial \ are \frac{2}{3} \ and \frac{-1}{4}}$...Given

$\bold{Let \alpha \ be \frac{2}{3} \ and \beta \ be \frac{-1}{4}}$

$\bold{\alpha+\beta=\frac{2}{3}+\frac{-1}{4}}$

$\bold{\alpha+\beta=\frac{5}{12}}$

$\bold{\alpha×\beta=\frac{2}{3}× \frac{-1}{4}}$

$\bold{\alpha×\beta=\frac{-2}{12}}$

Quadratic polynomial is

$\bold{x^{2}-(\alpha+\beta)x+(\alpha×\beta)}$

$\bold{x^{2}- \frac{5x}{12}- \frac{2}{12}}$

Verification of coefficients:-

In the above polynomial

a=1,b=$\bold{\frac{-5}{12}}$,c=$\bold{\frac{-2}{12}}$

$\bold{\alpha+\beta=\frac{-b}{a}}$

We know,$\bold{\alpha+\beta=\frac{5}{12}}$

Also

$\bold{\frac{-b}{a}=\frac{5}{12}}$

And

$\bold{\alpha×\beta=\frac{c}{a}}$

We know,$\bold{\alpha×\beta=\frac{-2}{12}}$

Also

$\bold{\frac{c}{a}=\frac{-2}{12}}$

Therefore required quadractic polynomial is

$\bold{x^{2}- \frac{5x}{12}- \frac{2}{12}}$.