Difference between revisions of "2013 UNCO Math Contest II Problems/Problem 4"
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== Solution == | == Solution == | ||
− | <math>x=1,-2,-5 | + | There are 3 cases in which <math>(x^2 - \tfrac{7}{2} x + \tfrac{3}{2} )^{(x^2+7x+10)}= 1</math>, |
+ | <ul> | ||
+ | <li><math>(x^2 - \tfrac{7}{2} x + \tfrac{3}{2}) = 1</math>.</li> | ||
+ | <cmath>\begin{align*} | ||
+ | x^2 - \tfrac72x + \tfrac32 &= 1 \ | ||
+ | x^2 - \tfrac72x + \tfrac12 &= 0 \ | ||
+ | x &= \dfrac{\tfrac72 \pm\sqrt{\tfrac{49}4-6}}{2} \ | ||
+ | x &= \dfrac{\tfrac72 \pm \tfrac{41}{2}}2 \ | ||
+ | x &= \dfrac{7\pm\sqrt{41}}{4} | ||
+ | \end{align*}</cmath> | ||
+ | <li><math>(x^2+7x+10) = 0</math>.</li> | ||
+ | <cmath>\begin{align*} | ||
+ | x^2 + 7x + 10 &= 0 \ | ||
+ | (x+5)(x+2) &= 0 \ | ||
+ | x &= -2, -5 | ||
+ | \end{align*}</cmath> | ||
+ | <li><math>(x^2 - \tfrac{7}{2} x + \tfrac{3}{2}) = -1</math>. | ||
+ | <cmath>\begin{align*} | ||
+ | x^2 - \tfrac72x + \tfrac32 &= -1 \ | ||
+ | x^2 - \tfrac72x + \tfrac52 &= 0 \ | ||
+ | x &= \dfrac{\tfrac72 \pm \sqrt{\tfrac{49}4-10}}{2} \ | ||
+ | x &= 1, \tfrac52 | ||
+ | \end{align*}</cmath> | ||
+ | However, note that since <math>(-1)</math> can be exponentiated to <math>-1</math>, we must check for extraneous solutions. | ||
+ | <cmath>\begin{align*} | ||
+ | (-1)^{1 + 7 + 10} &= (-1)^{18} \ | ||
+ | &= 1, | ||
+ | \end{align*}</cmath> | ||
+ | <cmath>\begin{align*} | ||
+ | (-1)^{\tfrac{25}4 + 7(\tfrac72) + 10} &= (-1)^{\tfrac{135}4} \ | ||
+ | &= (\sqrt[4]{-1})^{135} (\ne1). | ||
+ | \end{align*}</cmath> | ||
+ | </ul> | ||
+ | |||
+ | These cases give us <math>5</math> solutions: <math>\boxed{x=1,-2,-5,\tfrac{7\pm\sqrt{41}}{4}}</math>. | ||
+ | |||
+ | ~pineconee | ||
== See Also == | == See Also == | ||
{{UNCO Math Contest box|n=II|year=2013|num-b=3|num-a=5}} | {{UNCO Math Contest box|n=II|year=2013|num-b=3|num-a=5}} | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] |
Latest revision as of 13:58, 30 April 2022
Problem
Find all real numbers that satisfy
Solution
There are 3 cases in which ,
- .
- .
- . However, note that since can be exponentiated to , we must check for extraneous solutions.
These cases give us solutions: .
~pineconee
See Also
2013 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
All UNCO Math Contest Problems and Solutions |