Difference between revisions of "2023 WSMO Speed Round Problems/Problem 9"
(Created page with "==Problem== Suppose that <math>b</math> and <math>c</math> are the roots of the equation <math>x^2-\log(16)x+\log(64).</math> If <math>\sqrt{a+b}+\sqrt{a+c} = \sqrt{b+c},</ma...") |
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==Solution== | ==Solution== | ||
| + | We have | ||
| + | <cmath>\begin{align*} | ||
| + | \sqrt{a+b}+\sqrt{a+c} &= \sqrt{b+c}\implies\\ | ||
| + | \left(\sqrt{a+b}+\sqrt{a+c}\right)^2 &= \left(\sqrt{b+c}\right)^2\implies\\ | ||
| + | (a+b)+(a+c)+2\sqrt{(a+b)(a+c)} &= b+c\implies\\ | ||
| + | 2a+2\sqrt{(a+b)(a+c)} &= 0\implies\\ | ||
| + | 2\sqrt{(a+b)(a+c)} &= -2a\implies\\ | ||
| + | \left(\sqrt{(a+b)(a+c)}\right)^2 &= (-a)^2\implies\\ | ||
| + | (a+b)(a+c) &= a^2\implies\\ | ||
| + | a^2+ab+ac+bc &= a^2\implies\\ | ||
| + | ab+ac+bc &= 0\implies\\ | ||
| + | a &= -\frac{bc}{b+c}. | ||
| + | \end{align*}</cmath> | ||
| + | From Vieta's Formulas, we have | ||
| + | <cmath>\begin{align*} | ||
| + | a &= -\frac{\log(64)}{\log(16)}=-\log_{16}64 = -\log_{4^2}4^3=-\frac{3}{2}\implies\\ | ||
| + | 2^a &= 2^{-\frac{3}{2}} = \frac{1}{\sqrt{8}} = \frac{\sqrt{2}}{4}\implies 2+4 = \boxed{6}. | ||
| + | \end{align*}</cmath> | ||
| + | |||
| + | ~pinkpig | ||
Latest revision as of 11:12, 13 September 2025
Problem
Suppose that 
and 
are the roots of the equation 
If 
then 
find 
Solution
We have
From Vieta's Formulas, we have
~pinkpig
