Question

if zeroes of the quadratic polynomial x2+(a+1)x+b are 2 and -3 ,then find a=?, b=?

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Given:

quadratic polynomial: $x^2+(a+1)x+b$

zeroes of the polynomial: $2$ and $-3$

To find:

The values of a and b

Solution:

We know the general quadratic equation is:

$x^2-(\alpha +\beta )x+\alpha \beta =0$

and $ax^2+bx+c=0$

Now, $\alpha =2$ and $\beta =-3$.

So, we can compare the values for further calculations as follows,

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

⇒$\alpha +\beta =-\frac{(a+1)}{1}$

⇒$2-3=-(a+1)$

⇒$a+1=1$

⇒$a=0$

Also,

$\alpha \beta =\frac{c}{a}$

⇒$\alpha \beta =\frac{b}{1}$

⇒$2$×$(-3) =b$

⇒$b=-6$

Hence, the value of $a=0$ and $b=-6$.