Product of all the roots of the equation (5+26)x2−3+(5−26)x2−3=10 is

5−26=15+26, taking t=(5+26)x2−3Given equation bet+1t=10⇒t2−10t+1=0⇒t=5±26⇒t=(5+26)±1⇒x2−3=±1⇒x2=4 or 2⇒x=2,−2,2,−2Product of root=8

Product of all the roots of the equation (5+26)x2−3+(5−26)x2−3=10 is

5−26=15+26, taking t=(5+26)x2−3Given equation bet+1t=10⇒t2−10t+1=0⇒t=5±26⇒t=(5+26)±1⇒x2−3=±1⇒x2=4 or 2⇒x=2,−2,2,−2Product of root=8

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Product of all the roots of the equation $(5 + 2 \sqrt{6})^{x^{2} - 3} + (5 - 2 \sqrt{6})^{x^{2} - 3} = 10$ is

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answer is A.

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Detailed Solution

$5 - 2 \sqrt{6} = \frac{1}{5 + 2 \sqrt{6}}, taking t = (5 + 2 \sqrt{6})^{x^{2} - 3}$

Given equation be
$\begin{array}{l} t + \frac{1}{t} = 10 \Rightarrow t^{2} - 10 t + 1 = 0 \Rightarrow t = 5 \pm 2 \sqrt{6} \\ \Rightarrow t = (5 + 2 \sqrt{6})^{\pm 1} \\ \Rightarrow x^{2} - 3 = \pm 1 \Rightarrow x^{2} = 4 or 2 \Rightarrow x = 2, - 2, \sqrt{2}, - \sqrt{2} \end{array}$
Product of root=8