# ((x+6)/(x^(1/2))=35) ( (x+1)/(x)=?)

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#### Explanation

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#### Explanation:

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1
Feb 22, 2018

$\frac{x + 1}{x} = \frac{1285}{72} \pm \frac{35}{72} \sqrt{1201}$

#### Explanation:

Given:

$\frac{x + 6}{x} ^ \left(\frac{1}{2}\right) = 35$

Multiply both sides by ${x}^{\frac{1}{2}}$ to get:

$x + 6 = 35 {x}^{\frac{1}{2}}$

Square both sides to get:

${x}^{2} + 12 x + 36 = 1225 x$

Subtract $1225 x$ from both sides to get:

${x}^{2} - 1213 x + 36 = 0$

Next note that we want to find:

$\frac{x + 1}{x} = 1 + \frac{1}{x}$

Multiplying the quadratic we have found by $\frac{1}{x} ^ 2$ we get:

$36 {\left(\frac{1}{x}\right)}^{2} - 1213 \left(\frac{1}{x}\right) + 1 = 0$

So by the quadratic formula we find:

$\frac{1}{x} = \frac{1213 \pm \sqrt{{\left(- 1213\right)}^{2} - 4 \left(36\right) \left(1\right)}}{2 \left(36\right)}$

$\textcolor{w h i t e}{\frac{1}{x}} = \frac{1213 \pm \sqrt{1471369 - 144}}{72}$

$\textcolor{w h i t e}{\frac{1}{x}} = \frac{1213 \pm \sqrt{1471225}}{72}$

$\textcolor{w h i t e}{\frac{1}{x}} = \frac{1213 \pm 35 \sqrt{1201}}{72}$

So:

$\frac{x + 1}{x} = 1 + \frac{1}{x} = \frac{1285}{72} \pm \frac{35}{72} \sqrt{1201}$

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

1
Dani W. Share
Feb 22, 2018

1

#### Explanation:

Solve for x:

$\frac{x + 6}{x} ^ \left(\frac{1}{2}\right) = 35$
$x + 6 = 35 {x}^{\frac{1}{2}}$

I chose to square both sides in order to get rid of the square root.

${\left(x + 6\right)}^{2} = 1225 x$
${x}^{2} + 12 x + 36 = 1225 x$
${x}^{2} - 1213 x + 36 = 0$

I don't think I can factor this, so I'm going to apply the quadratic formula instead!

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
$x = \frac{1213 \pm 5 \sqrt{58849}}{2}$
$x = \frac{1213 + 5 \sqrt{58849}}{2}$ because $\frac{\left(\frac{1213 + 5 \sqrt{58849}}{2}\right) + 6}{\sqrt{\frac{1213 + 5 \sqrt{58849}}{2}}} = 35$

Now all you have to do is plug $x = \frac{1213 + 5 \sqrt{58849}}{2}$ into $\frac{x + 1}{x}$!

$\frac{x + 1}{x} \approx 1$

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