Question

find a quadratic polynomial if the zeroes if the zeroes of it are 2 and 1/3 respectively between zeroes and coefficients​

Answer 1

$\large\underline{\sf{Solution-}}$

Let f(x) be a required Quadratic polynomial having zeroes 2 and 1/3 respectively.

Let further assume that,

$\rm :\longmapsto\: \alpha = 2$

and

$\rm :\longmapsto\: \beta = \dfrac{1}{3}$

So,

➢Sum of zeroes of quadratic polynomial is

$\rm :\longmapsto\: \alpha + \beta$

$\rm \: = \: \: 2 + \dfrac{1}{3}$

$\rm \: = \: \: \dfrac{6 + 1}{3}$

$\rm \: = \: \: \dfrac{7}{3}$

$\bf\implies \: \alpha + \beta = \dfrac{7}{3}$

Now,

➢Product of zeroes of quadratic polynomial is

$\rm :\longmapsto\: \alpha \beta$

$\rm \: = \: \: 2 \times \dfrac{1}{3}$

$\rm \: = \: \: \dfrac{2}{3}$

$\bf\implies \: \alpha \beta = \dfrac{2}{3}$

We know that

➢Quadratic polynomial is given by

$\rm :\longmapsto\:f(x) = k\bigg( {x}^{2} - ( \alpha + \beta ) x+ \alpha \beta \bigg), \: where \: k \ne \: 0$

On substituting the values, we get

$\rm :\longmapsto\:f(x) = k\bigg( {x}^{2} - \dfrac{7}{3} x+ \dfrac{2}{3} \bigg), \: where \: k \ne \: 0$

$\rm :\longmapsto\:f(x) = \dfrac{k}{3} \bigg( 3{x}^{2} - 7x+ 2 \bigg), \: where \: k \ne \: 0$

$\bf\implies \:f(x) = {3x}^{2} - 7x + 2$

Verification :-

We know,

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\rm :\longmapsto\:\sf Sum\ of\ the\ zeroes= - \dfrac{( - 7)}{3} = \dfrac{7}{3}$

Hence, Verified

Also,

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\rm :\longmapsto\:\sf Product\ of\ the\ zeroes=\dfrac{2}{3}$

Hence, Verified.

Thus, we concluded that

$\boxed{\red{\sf \alpha + \beta =\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

and

$\boxed{\red{\sf \alpha \beta =\frac{Constant}{coefficient\ of\ x^{2}}}}$