Question

If $\left(7-4\sqrt{3}\right)^{x^2-4x+3} + \left(7+4\sqrt{3}\right)^{x^2-4x+3} = 14,$ then the value of x is given by

• A. $2, 2 \pm \sqrt{2}$
• B. $ \sqrt{2} \pm \sqrt{3} , 3$
• C. $3 \pm \sqrt{2} , 2$
• D. None of these

Answer 1

Answer: (A)

Since $\left(7-4\sqrt{3}\right) + \left(7+4\sqrt{3}\right)= 1$
$\therefore $ The given equation becomes
$y + \frac{1}{y} = 14$ where $y = \left(7-4\sqrt{3}\right)^{x^2-4x+3} $
$\Rightarrow y^{2 }- 14y + 1 = 0 \Rightarrow y = 7 \pm 4\sqrt{3}$
Now $y = 7 + 4\sqrt{3} \Rightarrow x^{2 }- 4x + 3 = -1 \Rightarrow x = 2, 2$
Also $y = 7 - 4 \sqrt{3} \Rightarrow x^{2 }- 4x + 3 = 1 \Rightarrow x = 2 \pm \sqrt{2}$