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 $y=7-4\sqrt{3}\Rightarrow x^2-4x+3=1\Rightarrow x=2\pm \sqrt{2}$