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There are two equations: $x^2+2p_1x + q_1 = 0$ and $x^2 + 2p_2x+q_2 = 0$. It is also known that $q_1+q_2 = 2p_1p_2$. How can I prove that if one of them doesn't have roots, then the other one has the roots?

So I have

$$4p_1^2 - 4q_1 < 0$$

I need to prove that $$p_2^2 - q_2 > 0$$.

I tried putting $2p_1p_2-q_1$ instead of $q_2$ but it gave me nothing. Can you please help me figure this problem out?

1 Answers 1

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Hint: $\;(p_1^2-q_1)+(p_2^2-q_2)=p_1^2+p_2^2-2p_1p_2=(p_1-p_2)^2 \ge 0\,$.

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    I think I figured it out. You combine the two discriminants and get this thing. We also know that D1 is negative. If we add a number to a negative one and it becomes positive that means D2 is positive. Right? – 2017-02-01
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    @idliketodothis Right. Otherwise put, the sum of the two discriminants is non-negative, so (at least) one of them must be non-negative (because if they were both negative, then their sum would be negative as well). – 2017-02-01