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If the roots of the equation $ax^2 + 2bx + c=0$ are real and disrinct, then show that the roots of the equation $x^2-2(a+b)x+a^2+b^2+2c^2=0$ are non-real complex numbers.

For the first equation discriminant >0, which gives $b^2-ac>0$.

for the second equation we have to show that discriminant <0. that is where I am facing problem.

$\Delta = 4(a+b)^2-4(a^2+b^2+2c^2)= 8(ab-c^2)$

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  • $\begingroup$ Compute the discriminant. $\endgroup$ – Wuestenfux May 31 at 7:29
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I think the second equation it's $$x^2-2(a+c)x+a^2+c^2+2b^2=0.$$ If so, your reasoning gives a proof.

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You can not show what you have to show, since, if $a=b=1$ and $c=0$, then both equations have real and distinct roots.

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