-1
$\begingroup$

If $p$ and $q$ are the roots of equation $x^2+mx+n=0$ and $m$ and $n$ are the roots of equation $x^2+px+q=0$ then the equation whose roots are $(q+n)$ and $(p+m)$ is

$A) \; x^2-4$

$B) \; x^2-2x+4 $

$C) \; x^2+4x-4 $

$D) \; x^2+2x-8 $

Note that $p, q, r$ and $s$ are non zero integers.

  • 0
    Have you tried anything, like using Vieta's formulas? – 2017-02-18
  • 0
    Yes but didn't get the answer please help.. – 2017-02-18
  • 0
    The answer is D. Does that help? – 2017-02-18
  • 0
    How did you get the answer.explain briefly ?? – 2017-02-18

2 Answers 2

1

You have $p+q=-m$ and $pq=n$ from the first equation and $m+n=-p$ and $mn=q$ from the second equation.

For the new quadratic equation, the sum of roots is $q+n+p+m=n=q$ using the above, so it follows that $p=1=m$, and therefore $q=-2=n$

Now calculate the new product of roots and you get $-8$ so the answer is D

I hope this helps

  • 0
    You are welcome – 2017-02-18
1

Using the Vieta's formulas we get:

\begin{array}{c c} p + q = -m, & pq = n \\ m + n = -p, & mn = q \end{array}

From there we get $m + n = -p = m + q\implies n = q$. Since $n,q\neq 0$, we have that $ p = m = 1$ and $q = n = -2$. So, $p + m = 2$, $q + n = -4$, and the answer is $(x-2)(x+4) = x^2 +2x -8$.