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