Question

if zeros of polynomial x square +( A + 1 )x +b are 2 and minus 3 then find the value of (a + b)

Answer 1

Answer :

The value of (a+b) is -6

Given :

The quadratic polynomial is :

• x² + (a + 1)x + b
• 2 and -3 are the zeroes of the polynomial

To Find :

• The value of (a + b)

Concept to be Used here :

If a polynomial ax² + bx + c has zeroes α and ß then :

• Sum of the zeroes = -b/a
• $\sf \implies \alpha + \beta = \dfrac{ - b}{a}$
• Product of the zeroes = c/a
• $\sf \implies \alpha \beta = \dfrac{c}{a}$

Solution :

Considering the given polynomial :

Sum of the zeroes = -coefficient of x/coefficient of x²

$\implies\sf{ 2 + (-3) = \dfrac{-(a + 1)}{1} }\\\\ \sf{\implies 2-3 = -a - 1}\\\\ \sf{\implies -a - 1= -1}\\\\ \implies\sf{a + 1 = 1}\\\\ \sf{\implies a = 0 ...........(1)}$

And

Product of the zeroes= constant term/coefficient of x²

$\implies\sf{(2)(-3) = \dfrac{b}{1}}\\\\ \sf{\implies -6 = b}\\\\ \sf{\implies b = -6 ............(2)}$

Adding (1) and (2) we have :

$\implies\sf{a+b = 0+(-6)}\\\\ \implies\bf{a+b= -6 }$

Answer 2

Given equation

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

2 and -3 are zeroes of the polynomial.

Substituting 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 -3 in the equation, we get,

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

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

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

Therefore,

a+b = -6