Question

$(7-4\sqrt{3}{)}^{x^2-4x+3}+(7+4\sqrt{3}{)}^{x^2-4x+3}=14$

Answer 1

$(7-4\sqrt{3})(7+4\sqrt{3})=7^2-(4\sqrt{3}{)}^2$

$=49-48=1$

$⟹7+4\sqrt{3}=\frac{1}{7-4\sqrt{3}}$

From the given equation, we can observed:

$(7-4\sqrt{3}{)}^{x^2-4x+3}+(7+4\sqrt{3}{)}^{x^2-4x+3}=14$

$(7-4\sqrt{3})+(7+4\sqrt{3})=7+7\dots 1$

On comparing the above equations, we get that:

$x^2-4x+3=1$ or $x^2-4x+3=-1$

$x^2+4x+2=0$ or $x^2-4x+4=0$

$\therefore x=-2\pm \sqrt{2}$ or $x=2,2$