# One root of the equation $x^2 + px + q = 0$ is $n$ times the other [closed]

One root of the equation $$x^2 + px + q = 0$$ is $$n$$ times the other, where $$n ≠ 0$$. Show that $$qn^2 + (2q – p^2)n + q = 0$$.

These are the answers. This is what I don't get... Someone explain to me how they got to this conclusion • Which of the four lines of the given answer bothers you?
– YNK
Mar 6, 2022 at 9:10
• I don't get the transition from 2nd line to the third line Mar 6, 2022 at 9:11
• Note that alternatively you can assume the roots $\alpha, n\alpha$ and then plug the Vieta's formulas $q=\alpha^2n$ and $p=-n\alpha-\alpha$ into the $qn^2+(2q-p^2)n+q$ to see it simplifies to $0$.
– Sil
Mar 6, 2022 at 9:24

The first line of the solution is a direct application of the quadratic formula.

If $$ax^2 + bx + c = 0$$ where $$a \neq 0$$, then $$x=\frac{-b\pm \sqrt{b^2 - 4ac}}{2a}$$.
The second line is because of the fact that one root is $$n$$ times the other.
From second to third line, expansion will do the job. $$-p + \sqrt{p^2 - 4q} = -np -n\sqrt{p^2 - 4q}$$ $$\sqrt{p^2 - 4q} + n\sqrt{p^2 - 4q}=-np+p$$ $$\sqrt{p^2 - 4q}= \frac{-np+p}{1+n}$$ We then square both sides of the equation. $$p^2 - 4q = \frac{(-np+p)^2}{(1+n)^2}$$ Now, multiply the denominator to the LHS, expand both brackets and group the terms in descending order of $$n$$. We shall get the answer.