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In below problem I've been trying to figure out how simply letting $y=\dfrac{x}{x+10}$ gives the quadratic with roots scaled by $\dfrac{1}{\alpha+10}$. I'm a bit clueless why it works.

My thoughts :
- To get a quadratic $g(x)$ whose roots are $k$ times bigger than $f(x)$, we can simply replace $x$ by $x/k$. This works because $f(\alpha) = 0 \implies g(k\alpha) = 0$, where $g(x) = f(x/k).$
- To get a quadratic $g(x)$ whose roots are $h$ bigger than $f(x)$, we can simply replace $x$ by $x-h$. This works because $f(\alpha) = 0 \implies g(\alpha+h) = 0$, where $g(x) = f(x-h).$

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AgentS
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    It's certainly odd that they give an answer with a common factor of $2$, but the solution given seems perfectly lucid. What's your objection to it? – Angina Seng Jul 22 '18 at 09:38
  • @LordSharktheUnknown I've no objection as such, I just want to understand how it works. I know how to form a quadratic whose roots are scaled by a constant factor. But here they're scaling by a variable, which I feel is throwing me off.. – AgentS Jul 22 '18 at 09:41
  • Btw thanks for responding @LordSharktheUnknown :) I guess I'm looking for some reasoning/explanation behind letting $y=x/(x+10)$ – AgentS Jul 22 '18 at 09:42
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    You have $y=\phi(x)$ for some function. Then $x=\phi^{-1}(y)$ and if $Q(x)=0$ then $Q(\phi^{-1}(y))=0$. All the examples here are of this form. – Angina Seng Jul 22 '18 at 09:45
  • Wow @LordSharktheUnknown your explanation goes straight to my head. I see clearly now they're just replacing $x$ by $\phi^{-1}(y)$. Using inverse notation did the magic. Thank you so much :D If you kindly post above comment as answer, I'd love to mark best answer :) – AgentS Jul 22 '18 at 09:49
  • Why not use Vieta? – Michael Hoppe Jul 22 '18 at 10:07
  • @MichaelHoppe I'm familiar with Vieta formulas, not that sure how to make use of them here. I'd love to see XD – AgentS Jul 22 '18 at 10:10
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    If the desired equation reads $x^2+px+q=0$ you know that $q=\frac{a}{a+10}\frac{b}{b+10}$. Now exploit that $ab=-2$ and $a+b=-13$. – Michael Hoppe Jul 22 '18 at 10:14
  • $(a+10)(b+10) = ab +10(a+b) + 100 $, ahh we got $q$ easily. Beautiful! Let me try $p$ next @MichaelHoppe – AgentS Jul 22 '18 at 10:17
  • $-p = \dfrac{a}{a+10} + \dfrac{b}{b+10}$. Numerator is $a(b+10) + b(a+10) = 2ab+10(a+b)$. So cool XD thanks @MichaelHoppe – AgentS Jul 22 '18 at 10:20
  • @rsadhvika You’re welcome. – Michael Hoppe Jul 22 '18 at 10:52

1 Answers1

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The roots of $$x^2-13x-2=0$$ satisfy $$a+b =13, ab=-2$$

The transformed equation is $$(x-\frac {a}{a-10})(x-\frac {b}{b-10}) $$

Upon simplification we get $$(ab-10(a+b)+100)x^2 -(2ab-10(a+b))x+ab=0$$

Substitution for a+b and ab gives us $$-32x^2+134x-2=0$$

Thus the correct answer is $$-32x^2+134x-2=0$$

Mohammad Riazi-Kermani
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