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  4. If a+b+c=0 and a, b, c are rational, then the roots of the equation (b+c-a) x2+(c+a-b) x+(a+b-c)=0 are

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Q. If $a+b+c=0$ and $a, b, c$ are rational, then the roots of the equation $(b+c-a) x^{2}+(c+a-b) x+(a+b-c)=0$ are

Complex Numbers and Quadratic Equations

Solution:

We have,
$ D=(c+a-b)^{2}-4(b+c-a)(a+b-c) $
$=(a+b+c-2 b)^{2}-4(a+b+c-2 a) $
$(a+b+c-2 c) $
$=(-2 b)^{2}-4(-2 a)(-2 c)=4\left(b^{2}-4 a c\right) $
$=4\left[(-a-c)^{2}-4 a c\right]=4(a-c)^{2} $
$=[2(a-c)]^{2}=$ perfect square
$\therefore $ Roots are rational