Question

Solve the following equations:
$7x-\frac{\sqrt{3x^2-8x+1}}{x}={(\frac{8}{\sqrt{x}}+\sqrt{x})}^2$.

Answer 1

$7x-\frac{\sqrt{3x^2-8x+1}}{x}={(\frac{8}{\sqrt{x}}+\sqrt{x})}^2\frac{7x^2-\sqrt{3x^2-8x+1}}{x}={\left(\frac{8+x}{\sqrt{x}}\right)}^2\frac{7x^2-\sqrt{3x^2-8x+1}}{x}=\frac{{(8+x)}^2}{x}{7x}^2-\sqrt{3x^2-8x+1}={(8+x)}^2(x\ne 0){7x}^2-\sqrt{3x^2-8x+1}=64+x^2+16x6x^2-16x+2-\sqrt{3x^2-8x+1}=662(3x^2-8x+1)-\sqrt{3x^2-8x+1}=662{\left(\sqrt{3x^2-8x+1}\right)}^2-\sqrt{3x^2-8x+1}=66$

Put $t=\sqrt{3x^2-8x+1}$

$2t^2-t=66$

$2t^2-t-66=0$

$2t^2-12t+11t-66=0$

$2t(t-6)+11(t-6)=0(2t+11)(t-6)=0t=6,-\frac{11}{2}$

$\sqrt{3x^2-8x+1}=6,-\frac{11}{2}$

$3x^2-8x+1=36,\frac{121}{4}$

$3x^2-8x+1=36$

$3x^2-8x-35=0$

$3x^2-15x+7x-35=0$

$3x(x-5)+7(x-5)=0(3x+7)(x-5)+0$

$\Rightarrow x=5,-\frac{7}{3}$

Also, $3x^2-8x+1=\frac{121}{4}$

$12x^2-32x+4=121$

$12x^2-32x-117=0$

$x=\frac{32\pm \sqrt{1024+5616}}{24}=\frac{32\pm \sqrt{6640}}{24}$

$x=\frac{32\pm 4\sqrt{415}}{24}$

$x=\frac{8\pm \sqrt{415}}{6}$

So the values of $x$ are $-\frac{7}{3},5,\frac{8\pm \sqrt{415}}{6}$