# How do you solve 2sqrtx +1= 6/sqrtx?

Jul 18, 2015

Reformulate as a quadratic with possible spurious solutions and solve to find $x = \frac{9}{4}$ (and spurious solution $x = 4$)

#### Explanation:

Multiply both sides by $\sqrt{x}$ to get:

$2 x + \sqrt{x} = 6$

Subtract $2 x$ from both sides to get:

$\sqrt{x} = 6 - 2 x$

Square both sides to get:

$x = {\left(6 - 2 x\right)}^{2} = 36 - 24 x + 4 {x}^{2}$

Note that squaring may introduce spurious solutions, so need to check later.

Subtract $x$ from both sides to get:

$4 {x}^{2} - 25 x + 36 = 0$

Use the quadratic formula to find:

$x = \frac{25 \pm \sqrt{{25}^{2} - \left(4 \times 4 \times 36\right)}}{2 \cdot 4}$

$= \frac{25 \pm \sqrt{625 - 576}}{8}$

$= \frac{25 \pm \sqrt{49}}{8}$

$= \frac{25 \pm 7}{8}$

That is $x = 4$ or $x = \frac{9}{4}$

If $x = 4$ then:

$2 \sqrt{x} + 1 = 2 \sqrt{4} + 1 = 4 + 1 = 5$

But $\frac{6}{\sqrt{x}} = \frac{6}{\sqrt{4}} = \frac{6}{2} = 3$

So $x = 4$ is not a solution of the original equation.

If $x = \frac{9}{4}$ then:

$2 \sqrt{x} + 1 = 2 \sqrt{\frac{9}{4}} + 1 = 2 \cdot \frac{3}{2} + 1 = 3 + 1 = 4$

and $\frac{6}{\sqrt{x}} = \frac{6}{\sqrt{\frac{9}{4}}} = \frac{6}{\frac{3}{2}} = \frac{12}{3} = 4$

So the solution of the original equation is $x = \frac{9}{4}$