(5+26)x2−3+(5−26)x2−3=10

We have,(5+26)x2−3+(5−26)x2−3=10.⇒(5+25)x2−3+15+26x2−3=10⇒(5+25)x2−3+1(5+26)x2−3=10⇒a+1a=10,a=(5+26)x2−3⇒a2−10a+1=0⇒a=10±100−42=10±462=5±26When a=5+26⇒ (5+26)x2−3=(5+26)⇒ x2−3=1⇒ x2=4⇒ x=±2When a=5−26⇒ (5+26)x2−3=(5−26)⇒ (5+26)x2−3=(5+26)−1⇒ x2−3=−1⇒ x2=2⇒ x=±2Hence the solution set is {−2,−2,2,2}

(5+26)x2−3+(5−26)x2−3=10

We have,(5+26)x2−3+(5−26)x2−3=10.⇒(5+25)x2−3+15+26x2−3=10⇒(5+25)x2−3+1(5+26)x2−3=10⇒a+1a=10,a=(5+26)x2−3⇒a2−10a+1=0⇒a=10±100−42=10±462=5±26When a=5+26⇒ (5+26)x2−3=(5+26)⇒ x2−3=1⇒ x2=4⇒ x=±2When a=5−26⇒ (5+26)x2−3=(5−26)⇒ (5+26)x2−3=(5+26)−1⇒ x2−3=−1⇒ x2=2⇒ x=±2Hence the solution set is {−2,−2,2,2}

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$(5 + 2 \sqrt{6})^{x^{2} - 3} + (5 - 2 \sqrt{6})^{x^{2} - 3} = 10$

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-2,2,-22

${- 2, - \sqrt{2}, \sqrt{2}, 2}$

$2, 2$

0,2

answer is B.

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Detailed Solution

We have,

$\begin{array}{l} (5 + 2 \sqrt{6})^{x^{2} - 3} + (5 - 2 \sqrt{6})^{x^{2} - 3} = 10 . \\ \Rightarrow (5 + 2 \sqrt{5})^{x^{2} - 3} + {(\frac{1}{5 + 2 \sqrt{6}})}^{x^{2} - 3} = 10 \\ \Rightarrow (5 + 2 \sqrt{5})^{x^{2} - 3} + \frac{1}{(5 + 2 \sqrt{6})^{x^{2} - 3}} = 10 \\ \Rightarrow a + \frac{1}{a} = 10, a = (5 + 2 \sqrt{6})^{x^{2} - 3} \\ \Rightarrow a^{2} - 10 a + 1 = 0 \\ \Rightarrow a = \frac{10 \pm \sqrt{100 - 4}}{2} \end{array}$

$= \frac{10 \pm 4 \sqrt{6}}{2} = 5 \pm 2 \sqrt{6}$

When $a = 5 + 2 \sqrt{6}$

$\begin{array}{l} \Rightarrow (5 + 2 \sqrt{6})^{x^{2} - 3} = (5 + 2 \sqrt{6}) \\ \Rightarrow x^{2} - 3 = 1 \\ \Rightarrow x^{2} = 4 \\ \Rightarrow x = \pm 2 \end{array}$

When $a = 5 - 2 \sqrt{6}$
$\Rightarrow (5 + 2 \sqrt{6})^{x^{2} - 3} = (5 - 2 \sqrt{6})$

$\begin{array}{l} \Rightarrow (5 + 2 \sqrt{6})^{x^{2} - 3} = (5 + 2 \sqrt{6})^{- 1} \\ \Rightarrow x^{2} - 3 = - 1 \\ \Rightarrow x^{2} = 2 \\ \Rightarrow x = \pm \sqrt{2} \end{array}$