Known that: $(7+4\sqrt 3)^{x^2-8}+(7-4\sqrt 3)^{x^2-8}=14$ What is the value of $x$
Find $x$ if $(7+4\sqrt 3)^{x^2-8}+(7-4\sqrt 3)^{x^2-8}=14$
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algebra-precalculus
3 Answers
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Let $a=(7+4\sqrt{3})^{x^2-8}$ then $\dfrac{1}{a}=(7-4\sqrt{3})^{x^2-8}$
So the given equation will be $a+\dfrac{1}{a}=14\implies a^2-14a+1=0$
Solving quadratic for $a=7+4\sqrt{3}$ or $a=7-4\sqrt{3}\implies x^2-8=1$ or $-1$
therefore possible values of $x=3,-3,\sqrt{7},-\sqrt{7}$
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Hint: if $a$ be the first term on LHS, then the second term is $\frac{1}{a}$.
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Note that $(7+4 \sqrt{3}) \cdot (7-4\sqrt{3}) = 49-48 =1$