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 answers

  • 1
    $\begingroup$ Welcome to MSE. Please read this text about how to ask a good question. $\endgroup$ Mar 6, 2022 at 8:57
  • $\begingroup$ Which of the four lines of the given answer bothers you? $\endgroup$
    – YNK
    Mar 6, 2022 at 9:10
  • $\begingroup$ I don't get the transition from 2nd line to the third line $\endgroup$
    – Woo Luke
    Mar 6, 2022 at 9:11
  • $\begingroup$ 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 $0$. $\endgroup$
    – Sil
    Mar 6, 2022 at 9:24

2 Answers 2


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

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.

  • $\begingroup$ Wow, that helped well! Thank you for expanding everything! $\endgroup$
    – Woo Luke
    Mar 6, 2022 at 9:18

Here, They use Sridhar Acharya's Rule to find out root of Quadratic equation.

After finding roots of given equation,use the condition given in your question & then simplify it. You will get required answer.


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