Sovling : $3x^{-1/2} - \frac{2\sqrt{x} + 4}{5x} = \frac{1}{\sqrt{x}}$

Question

How would I solve the following equation to find $x$? I am struggling to get the correct answer.

$$3x^{-1/2} - \frac{2\sqrt{x} + 4}{5x} = \frac{1}{\sqrt{x}}$$

Answer 1

Set $y = \sqrt x$. Then your equation becomes

$$\frac3y - \frac{2y+4}{5y^2} = \frac1y$$

Rearranging,

$$\frac2y - \frac{2}{5y} - \frac{4}{5y^2} = 0$$

$$\frac{8}{5y} = \frac{4}{5y^2}$$

$$\frac2y = \frac{1}{y^2}$$

Now $y\neq0$, otherwise these equations would be nonsense, so we can cancel $y$:

$$2 = \frac1y$$

$$y = \frac12$$

$$x = y^2 = \frac14$$

Check that this satisfies the original equation:

$$\frac{3}{\sqrt{1/4}} - \frac{2\sqrt{1/4}+4}{5\times{1/4}} = \frac{1}{\sqrt{1/4}}$$

Answer 2

First of all, multiply by $x\sqrt{x}$ to get rid of the denominators. Note that $x^{-1/2} = \frac{1}{\sqrt{x}}$, thus :

$$3x^{-1/2} - \frac{2\sqrt{x} + 4}{5x} = \frac{1}{\sqrt{x}} \Leftrightarrow3x - \frac{2x+4\sqrt{x}}{5}= x \Leftrightarrow15x - 2x - 4\sqrt{x}=5x$$

$$\Leftrightarrow$$

$$2x - \sqrt{x} = 0$$

HINT : Let $\sqrt{x} = y$. Now can you reform the final equation and solve it ?

Answer 3

Hint : Do the change of variable $X=x^{1/2}$ and multiply by $X$ in both sides of the equation.

Answer 4

Rewrite the equation as $$ 3\frac{1}{\sqrt{x}}-\frac{2}{5}\frac{1}{\sqrt{x}}-\frac{4}{5}\frac{1}{x}=\frac{1}{\sqrt{x}} $$ If you set $t=1/\sqrt{x}$, the equation becomes $$ 3t-\frac{2}{5}t-\frac{4}{5}t^2=t $$ or $$ 4t^2-8t=0 $$ The solution $t=0$ cannot be accepted, because it would lead to the contradictory $1/\sqrt{x}=0$. Only $t=2$ remains, thus $$ \frac{1}{\sqrt{x}}=2 $$ and $x=1/4$.

Answer 5

$3x^{-1/2} - \frac{2\sqrt{x} + 4}{5x} = \frac{1}{\sqrt{x}}$

$3\frac 1{\sqrt x} - \frac {2\sqrt{x} + 4}{5x} = \frac 1{\sqrt x}$. Multiply both sides by $\sqrt {x}$ to simplify. (We'll have to make a mental note to remember that $x > 0$)

$3 - \frac {2x + 4\sqrt x}{5x} = 1$.

$- \frac {2x + 4\sqrt x}{5x} = -2$

$\frac {2x + 4\sqrt x}{5x} = 2$ Multiply both side by $5x$.

$2x + 4\sqrt x = 2*5x = 10x$

$4\sqrt x = 8x$

$\sqrt x = 2x$. Divide both sides by $\sqrt x$

$1 = 2 \frac x{\sqrt x} = 2 \sqrt x$

$\sqrt x = \frac 12$. Square both sides.

$x = \frac 14$.

Verify: $(\frac 14)^{-\frac 12} = \frac 1{\frac 14}^{\frac 12} = \frac 1{\frac 12} = 2$.

And $\frac{2\sqrt{\frac 14} + 4}{5*\frac 14} = \frac {2*\frac 12 + 4}{\frac 54} = \frac {1 + 4}{\frac 54} = \frac 5{\frac 54} = \frac {4*5}5 = 4$.

So $3(\frac 14)^{-\frac 14} - \frac{2\sqrt{\frac 14} + 4}{5*\frac 14} = 3*2 -4 = 2$.

And $\frac 1{\sqrt {\frac 14}} = \frac 1{\frac 12} = 2$ so ...

indeed $3(\frac 14)^{-\frac 14} - \frac{2\sqrt{\frac 14} + 4}{5*\frac 14}=\frac 1{\sqrt {\frac 14}} = 2$.