Question

If 2 and – 3 are the zeroes of the polynomial x2 + (a + 1)x + b, then find the value of a and b.

Answer 1

Answer:

a = 0 , b = -6

Step-by-step explanation:

See in the attachment.

Answer 2

Given polynomial

${x}^{2} + (a + 1)x + b$

Given zeroes of polynomial

$\alpha = 2 \\ \beta = - 3$

To find

values of ' a' and ' b'

Solution

Comparing given quadratic polynomial with the standard form of quadratic polynomial I. e,

$p {x}^{2} + qx + r$

we will get ceffiecients

p = 1 ; q = a +1 ; r = b

Now,

as we know by the relation of zeroes and coeffients

that,

$\alpha + \beta = \frac{ - (q)}{p} \\ \\ (putting \: values) \\ 2 + ( - 3) = \frac{ - (a + 1)}{1} \\ \\ - 1 = - a - 1 \\ \\ a \: = 0$

also,

$\alpha \beta = \frac{r}{p} \\ \\ (putting \: values) \\ (2)( - 3) = \frac{b}{1} \\ \\ - 6 =b \\ \\ b = - 6$

SO THE VALUE OF a Is 0 AND VALUE OF b IS -6.