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Difference between revisions of "2013 UNCO Math Contest II Problems/Problem 4"

m (→‎Problem: didn’t use latex formatting)
 
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== Solution ==
 
== Solution ==
−
<math>x=1,-2,-5,\tfrac{7\pm\sqrt{41}}{4}</math>
+
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 $x$ that satisfy $(x^2 - \tfrac{7}{2} x + \tfrac{3}{2} )^{(x^2+7x+10)}= 1.$

Solution

There are 3 cases in which $(x^2 - \tfrac{7}{2} x + \tfrac{3}{2} )^{(x^2+7x+10)}= 1$,

  • $(x^2 - \tfrac{7}{2} x + \tfrac{3}{2}) = 1$.

    • \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*}

  • $(x^2+7x+10) = 0$.

    • \begin{align*} x^2 + 7x + 10 &= 0 \\ (x+5)(x+2) &= 0 \\ x &= -2, -5 \end{align*}

  • $(x^2 - \tfrac{7}{2} x + \tfrac{3}{2}) = -1$. \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*} However, note that since $(-1)$ can be exponentiated to $-1$, we must check for extraneous solutions. \begin{align*} (-1)^{1 + 7 + 10} &= (-1)^{18} \\ &= 1, \end{align*} \begin{align*} (-1)^{\tfrac{25}4 + 7(\tfrac72) + 10} &= (-1)^{\tfrac{135}4} \\ &= (\sqrt[4]{-1})^{135} (\ne1). \end{align*}

These cases give us $5$ solutions: $\boxed{x=1,-2,-5,\tfrac{7\pm\sqrt{41}}{4}}$.

~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
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