Problems need HardWork #5

$\begin{aligned} 2\sqrt{x} + \frac{2}{\sqrt{x}} & = 5 \\ \left(2\sqrt{x} + \frac{2}{\sqrt{x}} \right)^2 & = 5^2 \\ 4x + 8 + \frac{4}{x} & = 25 \\ 4x^2 - 17x + 4 & = 0 \\ 4x^2 - 17x & = -4 \\ \color{#3D99F6}{8}x^2 + \color{#D61F06}{(-34)}x & = -8 \end{aligned}$

$\Rightarrow \color{#3D99F6}{a} + \color{#D61F06}{b} = \color{#3D99F6}{8} \color{#D61F06}{-34} = \boxed{-26}$

$2\sqrt { x } +\frac { 2 }{ \sqrt { x } } =5\\ 2x+2=5\sqrt { x } \\ (2x+2)^{ 2 }=25x\\ 4{ x }^{ 2 }+8x+4=25x\\ 4{ x }^{ 2 }-17x=-4\\ 8{ x }^{ 2 }-34x=-8\\ a+b=8-34=-26$

First we need to find the roots of the first equation. $2\sqrt { x } +\frac { 2 }{ \sqrt { x } } =5\Leftrightarrow 2\sqrt { x } +\frac { 2 }{ \sqrt { x } } -5=0$ Let $y=\sqrt { x }$ then the equation turns into : $2y+\frac { 2 }{ y } -5=0\Leftrightarrow \frac { 2{ y }^{ 2 }-5y+2 }{ y } =0\Leftrightarrow 2{ y }^{ 2 }-5y+2=0$ .

From which we can conclude either $y=\frac{1}{2}$ or $y=2$ , which concludes $x=\frac{1}{4}$ or $x=4$ .

Then we move on to the next equation, $ax^2+bx=-8$ .

This equation can be turned into $ax^2+bx+8=0$ .

$ax^2+bx+8=0$ has roots of $x=\frac{1}{4}$ and $x=4$ , hence $ax^2+bx+8=a(x-\frac{1}{4})(x-4)=ax^2-a\frac { 17x }{ 4 } +a$ . Thus, $a=8$ and $-\frac { 17a }{ 4 } =b \Leftrightarrow b=-34$ .

In conclusion, the answer to the question is $8-34=\boxed{-26}$