If the zeroes of the quadratic polynomial $x^2+( a+1)x+b$ are $2$ and $-3$, then $a=?,\ b=?$.


Given: Quadratic polynomial $x^2+( a+1)x+b$ are $2$ and $-3$.

To do: To find the value of $a$ and $b$.

Solution:

$x^2+(a+1)x+b$ is the quadratic polynomial.

$2$ and $-3$ are the zeros of the quadratic polynomial.

Thus,

Sum of the zeroes$=2+(-3)=-\frac{a+1}{1}$
$\Rightarrow -\frac{a+1}{1}=-1$

$\Rightarrow a+1=1$

$\Rightarrow a=0$

Also, Product of the zeroes$=2\times (-3)=\frac{b}{1}$

$\Rightarrow b=-6$

Thus, $a=0,\ b=-6$.

Tutorialspoint
Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

15K+ Views

Kickstart Your Career

Get certified by completing the course

Get Started
Advertisements