Let S=x:x∈ℝ and (3+2)x2−4+(3−2)x2−4=10. Then n(S) is equal to

Let (3+2)x2−4=t given equation becomes t+1t=10 ⇒t2−10t+1=0⇒t=5+26,5−26 If (3+2)x2−4=5+26=(3+2)2⇒x=±6If (3+2)x2−4=5−26=(3+2)−2⇒x=±2

Let S=x:x∈ℝ and (3+2)x2−4+(3−2)x2−4=10. Then n(S) is equal to

Let (3+2)x2−4=t given equation becomes t+1t=10 ⇒t2−10t+1=0⇒t=5+26,5−26 If (3+2)x2−4=5+26=(3+2)2⇒x=±6If (3+2)x2−4=5−26=(3+2)−2⇒x=±2

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Let $S = \{x : x \in ℝ and (\sqrt{3} + \sqrt{2})^{x^{2} - 4} + (\sqrt{3} - \sqrt{2})^{x^{2} - 4} = 10\}$ . Then n(S) is equal to

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Detailed Solution

Let
$(\sqrt{3} + \sqrt{2})^{x^{2} - 4} = t \begin{array}{l} given equation becomes t + \frac{1}{t} = 10 \Rightarrow t^{2} - 10 t + 1 = 0 \\ \Rightarrow t = 5 + 2 \sqrt{6}, 5 - 2 \sqrt{6} \end{array}$
If $(\sqrt{3} + \sqrt{2})^{x^{2} - 4} = 5 + 2 \sqrt{6} = (\sqrt{3} + \sqrt{2})^{2}$
$\Rightarrow x = \pm \sqrt{6}$
If $(\sqrt{3} + \sqrt{2})^{x^{2} - 4} = 5 - 2 \sqrt{6} = (\sqrt{3} + \sqrt{2})^{- 2}$
$\Rightarrow x = \pm \sqrt{2}$