How do you solve $2 \sqrt{x} + 1 = \frac{6}{\sqrt{x}}$ $2sqrtx +1= 6/sqrtx$ ?

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How do you solve $2 \sqrt{x} + 1 = \frac{6}{\sqrt{x}}$ $2sqrtx +1= 6/sqrtx$ ?

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Answer 1 · 2015-07-18T10:32:54

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

Explanation:

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

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

Subtract $2 x$ $2x$ from both sides to get:

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

Square both sides to get:

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

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

Subtract $x$ $x$ from both sides to get:

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

Use the quadratic formula to find:

$x = \frac{25 \pm \sqrt{25^{2} - (4 \times 4 \times 36)}}{2 \cdot 4}$ $x = (25+-sqrt(25^2-(4xx4xx36)))/(2*4)$

$= \frac{25 \pm \sqrt{625 - 576}}{8}$ $= (25+-sqrt(625 - 576))/8$

$= \frac{25 \pm \sqrt{49}}{8}$ $= (25+-sqrt(49))/8$

$= \frac{25 \pm 7}{8}$ $= (25+-7)/8$

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

If $x = 4$ $x=4$ then:

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

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

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

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

$2 \sqrt{x} + 1 = 2 \sqrt{\frac{9}{4}} + 1 = 2 \cdot \frac{3}{2} + 1 = 3 + 1 = 4$ $2sqrt(x) + 1 = 2sqrt(9/4) + 1 = 2*3/2+1 = 3 + 1 = 4$

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

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

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How do you solve $2 \sqrt{x} + 1 = \frac{6}{\sqrt{x}}$ $2sqrtx +1= 6/sqrtx$ ?

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