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If x^2+a x+b=0a n dx^2+b x+c a=0(a!=b) h...

If `x^2+a x+b=0a n dx^2+b x+c a=0(a!=b)` have a common root, then prove that their other roots satisfy the equation `x^2+c x+a b=0.`

Text Solution

Verified by Experts

The correct Answer is:
`(bc_(1) - ab_(1)) (cb_(1) - ba_(1)) = aa_(1) - cc_(1))^(2)`
Subtracting the given equations, we get
` (a - b) x + c (b - c) = 0 rArr x = c ` is the common root
Thus, roots of `x^(2) + ax + bc = 0` are b and c and those of `x^(2) + bx + ca = 0`
are c and a . Also,` a + b = -c`. Thus, the required equation is
`x^(2) - (a + b) x + ab = 0`
`rArr x^(2) + cx + ab = 0` .
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Knowledge Check

  • If the equation x^(2) + bx + ca = 0 and x^(2) + cx + ab = 0 have a common root and b ne c , then their other roots will satisfy the equation

    A
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    A
    0
    B
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    `-1`
    D
    none of these

  • If the equations x^(2)-ax +b=0 and x^(2)+bx-a=0 have a common root, then

    A
    a=b
    B
    a+b=0
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