Show that that if $p,q,r,s$ are real numbers and $pr=2(q+s)$, then at least one of the eqns $x^2+px+q=0$ and $x^2+rx+s=0$ has real roots.

Question

My Attempt to the solution

we know to have a real solution d>=0 so either

1) $p^2-4q>=0$

or

2) $r^2-4s>=0$

or both are true.

Rearranging we get $(pr)^2 \geq 16qs$ substituting it in the first eqn we get $16qs\geq4q^2 +4s^2 +8qs$ So we get $0\geq(q-s)^2$ so we get $q=s$. Now what to do..? is there any another way to solve this?

If only one of the inequalities is guaranteed to be true, then you cannot combine them in any way. — Shaun Ault, Jul 05 '13 at 14:49
Assume that neither equation has real roots and start from there. It's not hard to deduce that then $pr \neq 2(q+s)$. — Daniel Fischer, Jul 05 '13 at 14:50
Moreover, the logic is backwards... we want to show that one of those ineqs is true, not start assuming one or both are true. — Shaun Ault, Jul 05 '13 at 14:50
Do as @DanielFischer said and consider the average of the discriminants. — S.B., Jul 05 '13 at 14:51

Shaun Ault · Accepted Answer · 2013-10-29T11:19:40.070

2

Given: $pr = 2(q+s)$. To prove: either $p^2 - 4q \geq 0$ or $r^2 - 4s \geq 0$. It is best to argue by contradiction -- assume both $p^2 - 4q < 0$ and $r^2 - 4s < 0$. Then upon rearranging and adding the inequalities, $$ \begin{array}{rcl} p^2 + q^2 &<& 4r + 4s \\ \frac{p^2 + q^2}{2} &<& 2(r + s) \end{array} $$ Can you see how to finish from here? If not, post a comment and I'll write more.

Hope this helps!

edited Oct 29 '13 at 11:19

answered Jul 05 '13 at 14:54

Shaun Ault

8,871

thnx. we will use am gm at the end.. – maths lover Jul 06 '13 at 01:36
3

There is a little correction in the above instead of $r$ the author wrote $q$. Given: $pr=2(q+s)$. To prove: either $p^2−4q≥0$ or $r^2−4s≥0$. It is best to argue by contradiction -- assume both $p^2−4q<0$ and $r^2−4s<0$. Then upon rearranging and adding the inequalities, $p^2+r^2 < 4q + 4s$ $\frac{p^2+r^2}{2} < 2(q + s)$ $\frac{p^2+r^2}{2} < pr$ $(pr=2(q + s))$ $(p^2+r^2) < 2pr$ $p^2 + r^2 - 2pr > 0$ $(p - r)^2 > 0$ it means $p = r$. – Oct 28 '13 at 08:20
Thank you @VIPINSETIA, I'm sorry I didn't see that obvious typo before! – Shaun Ault Oct 29 '13 at 11:20

Show that that if $p,q,r,s$ are real numbers and $pr=2(q+s)$, then at least one of the eqns $x^2+px+q=0$ and $x^2+rx+s=0$ has real roots.

1 Answers1