Question

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

Answer 1

2016 is anwer please mark as brain list you will also check in calculater

Answer 2

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:2 \: and \: - 3 \: are \: zeroes \: of \: {x}^{2} + (a + 1)x + b$

We know,

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

So,

$\rm :\longmapsto\: 2 + ( - 3) = - \dfrac{(a + 1)}{1}$

$\rm :\longmapsto\:2 - 3 = - a - 1$

$\rm :\longmapsto\: - 1 = - a - 1$

$\rm :\longmapsto\: - a = - 1 + 1$

$\rm :\longmapsto\: - a = 0$

$\bf\implies \:a = 0$

Also, we know that,

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\rm :\longmapsto\:(2) \times ( - 3) = \dfrac{b}{1}$

$\bf\implies \:b = - 6$

Alternative Method

Given that

$\rm :\longmapsto\:2 \: and \: - 3 \: are \: zeroes \: of \: {x}^{2} + (a + 1)x + b$

Let assume that

$\rm :\longmapsto\:f(x) \: = \: {x}^{2} + (a + 1)x + b$

We know,

Factor theorem states that, if a is a zero of polynomial f(x), then f( a ) = 0

So, using factor theorem, we have

$\rm :\longmapsto\:f(2) = 0$

$\rm :\longmapsto\: \: {2}^{2} +2 (a + 1) + b = 0$

$\rm :\longmapsto\: \: 4 +2a +2 + b = 0$

$\rm :\longmapsto\: \: 2a + b + 6 = 0$

$\bf :\longmapsto\: \: b = - 6 - 2a - - - (1)$

Also,

$\rm :\longmapsto\:f( - 3) = 0$

$\rm :\longmapsto\: \: {( - 3)}^{2} - 3 (a + 1) + b = 0$

$\rm :\longmapsto\: \: 9 - 3a - 3 + b = 0$

$\rm :\longmapsto\: \: 6 - 3a + b = 0$

$\rm :\longmapsto\: \: 6 - 3a - 6 - 2a= 0$

[ Using equation (1) ]

$\rm :\longmapsto\: - 5a = 0$

$\bf\implies \:a = 0$

Now, Substitute the value of a in equation (2), we get

$\rm :\longmapsto\:b = - 6 - 2(0)$

$\rm :\longmapsto\:b = - 6 -0$

$\bf\implies \:b = - 6$