Question

If the zeroes of the quadratic polynomial x2

+ (a + 1) x + b are 2 and -3, then

find the value of a and b​

Answer 1

Answer:

$a = 0$

$b = - 6$

I will tell you how it came.

Step-by-step explanation:

In the given quadratic polynomial

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

The zeroes are 2, -3

In px^2+qx+r let the zeroes be m&n.

Sum of zeroes = m+n = -b/a

product of zeroes = m*n = c/a

From the above formula,

p = 1,

q = (a+1) ,

r = b,

sum of zeroes =

$2 + ( - 3) = \frac{ - (a + 1)}{1} \\ - 1 = \frac{ - 1(a + 1)}{1} \\ a + 1 = 1 \\ a = 1 - 1 \\ a = 0$

Product of zeroes =

$2( - 3) = \frac{b}{1} \\ b = 2 \times - 3 \\ b = - 6$

If you like the process please comment