Question

the difference and the product of zeroes of a quadratic polynomial are 3 and 28 respectively and sum of its zeroes is a negative integer. find such a quadratic polynomial

Answer 1

Answer :

Let , Roots are $\alpha$ and $\beta$

Given , The difference of zeroes of a quadratic polynomial = 3

So,
$\alpha$ - $\beta$ = 3 ------ ( 1 )

Taking whole square on both hand side of equation 1 , we get

( $\alpha$ - $\beta$ )² = ( 3 )²

$$\begin{gathered} \Rightarrow {\alpha }^2+{\beta }^2-2\alpha \beta =9 \\ \Rightarrow {\alpha }^2+{\beta }^2-2\alpha \beta +2\alpha \beta -2\alpha \beta =9 \\ \Rightarrow {\alpha }^2+{\beta }^2+2\alpha \beta -4\alpha \beta =9 \\ \Rightarrow {\left(\alpha +\beta \right)}^2-4\alpha \beta =9----(2) \end{gathered}$$

And , The product of zeroes of a quadratic polynomial = 28

So,
$\alpha$ $\beta$ = 28 , Substitute that value in equation 2 , we get

$$\begin{gathered} \Rightarrow {\left(\alpha +\beta \right)}^2-4\left(28\right)=9 \\ \Rightarrow {\left(\alpha +\beta \right)}^2-112=9 \\ \Rightarrow {\left(\alpha +\beta \right)}^2=121 \\ \mathbf{\Rightarrow }\mathit{\alpha }\mathbf{}\mathbf{+}\mathbf{}\mathit{\beta }\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{\pm }\mathbf{11} \end{gathered}$$

Also given , sum of its zeroes is a negative integer , So

$\alpha$ + $\beta$ = - 11

And we know formula for polynomial when some of zeros and product of zeros we know :

Polynomial = k [ x² - ( Sum of zeros ) x + ( Product of zeros ) ] , Here k is any non zero real number.

Substitute values , we get

Quadratic polynomial = $k\left[x^2-\left(-11\right)x+28\right]=k\left[x^2+11x+28\right]$


= x² + 11 x + 28 [taking k = 1] ( Ans )