If the two equation x^2+ax+b=0 and x^2+bx+a=0 have a common root,show that other roots are the roots of the equation x^2+x+ab=0?

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Answer 1 · 2018-01-12T10:20:00

Proved.

Let The roots of the equation $x^2 + ax + b = 0$ be $alpha$ and $beta$ .

Similarly, Let the roots of the equation $x^2 + bx + a = 0$ be $alpha$ and $gamma$ . [There is a common root.]

So, $alpha + beta$ = $-a rArr alpha = -(a + beta)$

And $alpha + gamma$ = $-b rArr alpha = -(b + gamma)$

So, $-(a + beta) = -(b + gamma)$

$rArr a + beta = b + gamma$

$rArr beta - gamma = b -a$

$rArr (beta - gamma)^2 = b^2 - 2ab + a^2$ ...............(i)

Again, $alphabeta = b rArr beta = b/alpha$

And, $alphagamma = a rArr gamma = a/alpha$

Then, $beta - gamma = (b -a)/alpha$ and $betagamma = (ab)/alpha^2$

Putting this in eq (i)

$((b-a)/alpha)^2 = (b-a)^2$

$rArr 1/alpha^2 = 1$

$rArr alpha = +-1$

So, $betagamma = +-ab$ and $beta + gamma = (b + a)/alpha = +-(a + b)$

So, The Required Equation, which has $beta and gamma$ as roots is

$x^2 +- (a + b)x +- ab = 0$

But $(a + b) = +- 1$ [As, $alpha = +-1, beta = b$ and $gamma = a$ ]

So, The equation becomes $x^2 +-x +-ab = 0$

Hence Proved.