Solve for x 1/( square root of x+5) = square root of -2x

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

Step 1

Cross multiply.

Cross multiply by setting the product of the numerator of the right side and the denominator of the left side equal to the product of the numerator of the left side and the denominator of the right side.

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

Simplify the left side.

Combine using the product rule for radicals.

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

Step 2

To remove the radical on the left side of the equation, square both sides of the equation.

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

Step 3

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{- 2 x (x + 5)}$ as ${(- 2 x (x + 5))}^{\frac{1}{2}}$ .

${({(- 2 x (x + 5))}^{\frac{1}{2}})}^{2} = 1^{2}$

Simplify the left side.

Simplify ${({(- 2 x (x + 5))}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(- 2 x (x + 5))}^{\frac{1}{2}})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

${(- 2 x (x + 5))}^{\frac{1}{2} \cdot 2} = 1^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(- 2 x (x + 5))}^{\frac{1}{2} \cdot 2} = 1^{2}$

Rewrite the expression.

${(- 2 x (x + 5))}^{1} = 1^{2}$

Apply the distributive property.

${(- 2 x \cdot x - 2 x \cdot 5)}^{1} = 1^{2}$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

${(- 2 (x \cdot x) - 2 x \cdot 5)}^{1} = 1^{2}$

Multiply $x$ by $x$ .

${(- 2 x^{2} - 2 x \cdot 5)}^{1} = 1^{2}$

Multiply $5$ by $- 2$ .

${(- 2 x^{2} - 10 x)}^{1} = 1^{2}$

Simplify.

$- 2 x^{2} - 10 x = 1^{2}$

Simplify the right side.

One to any power is one.

$- 2 x^{2} - 10 x = 1$

Step 4

Solve for $x$ .

Subtract $1$ from both sides of the equation.

$- 2 x^{2} - 10 x - 1 = 0$

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = - 2$ , $b = - 10$ , and $c = - 1$ into the quadratic formula and solve for $x$ .

$\frac{10 \pm \sqrt{{(- 10)}^{2} - 4 \cdot (- 2 \cdot - 1)}}{2 \cdot - 2}$

Simplify.

Simplify the numerator.

Raise $- 10$ to the power of $2$ .

$x = \frac{10 \pm \sqrt{100 - 4 \cdot - 2 \cdot - 1}}{2 \cdot - 2}$

Multiply $- 4 \cdot - 2 \cdot - 1$ .

Multiply $- 4$ by $- 2$ .

$x = \frac{10 \pm \sqrt{100 + 8 \cdot - 1}}{2 \cdot - 2}$

Multiply $8$ by $- 1$ .

$x = \frac{10 \pm \sqrt{100 - 8}}{2 \cdot - 2}$

Subtract $8$ from $100$ .

$x = \frac{10 \pm \sqrt{92}}{2 \cdot - 2}$

Rewrite $92$ as $2^{2} \cdot 23$ .

Factor $4$ out of $92$ .

$x = \frac{10 \pm \sqrt{4 (23)}}{2 \cdot - 2}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{10 \pm \sqrt{2^{2} \cdot 23}}{2 \cdot - 2}$

Pull terms out from under the radical.

$x = \frac{10 \pm 2 \sqrt{23}}{2 \cdot - 2}$

Multiply $2$ by $- 2$ .

$x = \frac{10 \pm 2 \sqrt{23}}{- 4}$

Simplify $\frac{10 \pm 2 \sqrt{23}}{- 4}$ .

$x = \frac{5 \pm \sqrt{23}}{- 2}$

Move the negative in front of the fraction.

$x = - \frac{5 \pm \sqrt{23}}{2}$

The final answer is the combination of both solutions.

$x = - \frac{5 + \sqrt{23}}{2}, - \frac{5 - \sqrt{23}}{2}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{5 + \sqrt{23}}{2}, - \frac{5 - \sqrt{23}}{2}$

Decimal Form:

$x = - 4.89791576 \dots, - 0.10208423 \dots$

Solve for x 1/( square root of x+5) = square root of -2x

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