2013 UNCO Math Contest II Problems/Problem 4

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

2013 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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2013 UNCO Math Contest II Problems/Problem 4

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