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The question is:
Given that $m$ is a real number not less than $-1$, such that the equation in $x$ is $x^2+2(m-2)x+m^2-3m+3=0$ has two distinct roots $r$ and $s$. If $r^2+s^2=6$, find the value of $m$.

Here's what I've tried:.

Using Vieta's formulas:

$rs = m^2-3m+3$
$r+s = -2m+4$

Then I set the equation as such; $(r+s)^2 - 2rs = 6$. I got $2m^2-10m+4=0$. Using the quadratic equation, I got $\frac {5+\sqrt {17}}2$ and $\frac {5-\sqrt {17}}2$. Is there anything wrong with my solution?

GoodDeeds
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Isaiah Leobrera
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2 Answers2

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It is correct: $$(r+s)^2 - 2rs = 6 \iff \\ (-2(m-2))^2-2(m^2-3m+3)=6 \iff \\ (4m^2-16m+16)-2m^2+6m-6=6 \iff \\ 2m^2-10m+4=0.$$ And the roots are: $$m_{1,2}=\frac{5\pm \sqrt{17}}{2}.$$

farruhota
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The posted solution is correct up to that point, though it remains to be verified that $\,r \ne s\,$ for those values of $\,m\,$.

Alt. hint:   $\color{blue}{r+s}=-2m+4$, and each root satisfies $\,x^2= -\big(2(m-2)x+m^2-3m+3\big)\,$, so:

$$ \begin{align} 6 = r^2+s^2 &= -2(m-2)(\color{blue}{r+s})- 2(m^2-3m+3) \\ &= 2(m-2)(2m-4)- 2(m^2-3m+3) \\ &= 2 (m^2 - 5 m + 5) \end{align} $$

Then what remains to solve is $\,m^2-5m+5=3\,$ for solutions $\,\ge -1\,$ (and, again, verify that they yield distinct roots $\,r,s\,$).

dxiv
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  • good point about verifying, +1. Indeed, distinct roots imply $D>0 \iff (m-2)^2-m^2+3m-3>0 \iff m<1$. So, $\frac{5+\sqrt{17}}{2}>1$ must be rejected. – farruhota Sep 08 '18 at 08:00