Question

If a 2 and 3 are zero of x² + (a + 1 ) x + b then find 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,

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

Substituting x = 3, in the equation, we get,

$\implies (3)^{2} +(a+1)3+b=0\\\implies 9+3a+3+b=0\\\implies 3a+b+12=0\\\implies 3a+b=-12--(2)$

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

$2a+b=-6\\\underline{3a+b=-12}\\\underline{\underline{-a=6}}\\\implies a = -6$

Substituting a = -6 in the first equation, we get,

$\implies 2(-6)+b=-6\\\implies -12+b=-6\\\implies b = -6+12\\\implies b = 6$

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

                                                       

Answer 2

Correct Question:

If 2 and 3 are zeroes of $\sf{x^{2}+(a+1)x+b}$ then find the value of a and b.

Answer:-

The values of a and b are -6 and 6 respectively.

Given:

• The given quadratic polynomial is $\sf{x^{2}+(a+1)x+b}$

• Zeroes of the polynomial are 2 and 3.

To find:

• The value of a and b.

Solution:-

The given quadratic polynomial is $\sf{x^{2}+(a+1)x+b}$

Here a=1, b=a+1 and c=b

Sum of zeroes=$\sf{\frac{-b}{a}}$

$\sf{\therefore}$5=-(a+1)

$\sf{\therefore}$ -a-1=5

$\sf{\therefore}$ a=-6

Product of zeroes=$\sf{\frac{c}{a}}$

$\sf{\therefore}$ 6=b

$\sf{\therefore}$ b=6

The values of a and b are -6 and 6 respectively.