Question

If a and b are unequal and x2 +ax +band x2+ bx +a have a common factor the slow that a+ b + 1 = 0 .

Answer 1

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

Given that,

$\rm :\longmapsto\: {x}^{2} + ax + b \: and \: {x}^{2} + bx + a \: have \: a \: common \: factor.$

Let assume that

$\rm :\longmapsto\:p(x) = {x}^{2} + ax + b$

and

$\rm :\longmapsto\:q(x) = {x}^{2} + bx + a$

And further assume that,

$\rm :\longmapsto\:x - y \: is \: common \: factor \: of \: p(x) \: and \: q(x)$

We know,

Factor theorem :-

This theorem states that if x - a is a factor of polynomial f(x), then f(a) = 0.

So, By factor theorem, we have

$\rm :\longmapsto\:\boxed{ \tt{ \: p(y) = 0 \: \: \: and \: \: \: q(y) = 0 \: \: }}$

So,

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

and

$\rm :\longmapsto\: {y}^{2} + by + a = 0 - - - (2)$

On Subtracting equation (2) from equation (1), we get

$\rm :\longmapsto\: {y}^{2} + ay + b - {y}^{2} - by - a = 0$

$\rm :\longmapsto\: ay + b - by - a = 0$

$\rm :\longmapsto\: ay - by = a - b$

$\rm :\longmapsto\: (a- b)y = a - b$

$\bf\implies \:y = 1$

On substituting y = 1, in equation (1), we get

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

$\rm :\longmapsto\: 1 + a+ b = 0$

Hence,

$\bf\implies \:\boxed{ \tt{ \: a \: + \: b \: + \: 1 \: = \: 0 \: \: }}$