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

Given equation

$x^{2} +(a+1)x+b=0$

2 and -3 are zeroes of the polynomial.

Substituting x = 2 in the equation, we get,

$x^{2} +(a+1)x+b=0$

$\implies (2)^{2} +(a+1)2+b=0\\\implies 4+2a+2+b=0\\\implies 2a+b+6=0\\\implies 2a+b=-6--(1)$

Substituting x = 3 in the equation, we get,

$x^{2} +(a+1)x+b=0$

$\implies (-3)^{2} +(a+1)-3+b=0\\\implies 9-3a-3+b=0\\\implies -3a+b+6=0\\\implies -3a+b=-6--(2)$

Solving (1) and (2), we get,

$2a+b=-6\\\underline{-3a+b=-6}\\\underline{\underline{5a=0}}\\\implies a = 0\\\implies b = -6$

$\boxed{\boxed{\bold{Therefore, \ the \ values \ of \ a \ and \ b \ are \ 0 \ and \ -6 \ respectively}}}}}}}}$

                                                                     

Answer 2

Solution :

Given :

We have two zeroes p(x) = 2 and -3.

To find :

The value of a and b.

Explanation :

Putting the given equation,we get;

p(x) = 2

$\mapsto\tt{(2)^{2} +(a+1)2+b=0}\\\\\mapsto\tt{4+2a+2+b=0}\\\\\mapsto\tt{2a+b+6=0}\\\\\mapsto\tt{2a+b=-6....................(1)}$

&

p(x) = -3

$\mapsto\tt{(-3)^{2} +(a+1)-3+b=0}\\\\\mapsto\tt{9-3a-3+b=0}\\\\\mapsto\tt{-3a+b+6=0}\\\\\mapsto\tt{-3a+b=-6....................(2)}$

Using substitution method :

From eq. (1), we get;

$\mapsto\tt{b=-6-2a................(3)}$

Putting the value of b in equation (2),we get;

$\mapsto\tt{-3a+(-6-2a)=-6}\\\\\mapsto\tt{-3a-6-2a=-6}\\\\\mapsto\tt{-5a=-6+6}\\\\\mapsto\tt{-5a=0}\\\\\mapsto\tt{a=0/-5}\\\\\mapsto\bf{a=0}$

Putting the avlue of a in eq.(3),we get;

$\mapsto\tt{b=-6-2(0)}\\\\\mapsto\tt{b=-6-0}\\\\\mapsto\bf{b=-6}$

Thus;

The value of a and b will be 0 and -6 .