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Show that the roots of $ax^2+(a+b)x+b=0$ are real for all values of a and b.

I know that for the roots of the equation to be real the discriminant must be greater than zero.

So I've done $b^2-4ac=(a+b)^2-4ab $ and then expanded to $a^2+b^2-2ab$. But I don't see how this gives me a positive discriminant.

This is where I'm stumped, and don't know how to progress.

I apologise for any sloppy formatting, as this is my first time using mathematics exchange.

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    $\begingroup$ $ a^2 -2ab +b^2 = (a-b)^2 \geq 0$ $\endgroup$ – Nicholas Nov 12 '15 at 14:06
  • $\begingroup$ This is not the case unless $a$ and $b$ are constrained in some way - e.g. to be real numbers themselves or to be integers. One root will always be $x=-1$, but if $a$ and $b$ can be complex, the other root need not be real. $\endgroup$ – Mark Bennet Nov 12 '15 at 14:17
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We have $$ax^2+(a+b)x + b = ax^2+ax + bx+b = ax(x+1) + b(x+1) = (x+1)(ax+b)$$ Hence, $x=-1$ and $-b/a$ are the roots.


Proceeding your way, we have $a^2+b^2-2ab = (a-b)^2$, which is a non-negative discriminant.

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