Solve for x square root of x+4=x

$\sqrt{x + 4} = x$

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

${\sqrt{x + 4}}^{2} = x^{2}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x + 4}$ as ${(x + 4)}^{\frac{1}{2}}$ .

${({(x + 4)}^{\frac{1}{2}})}^{2} = x^{2}$

Simplify the left side.

Simplify ${({(x + 4)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(x + 4)}^{\frac{1}{2}})}^{2}$ .

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

${(x + 4)}^{\frac{1}{2} \cdot 2} = x^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(x + 4)}^{\frac{1}{2} \cdot 2} = x^{2}$

Rewrite the expression.

${(x + 4)}^{1} = x^{2}$

Simplify.

$x + 4 = x^{2}$

Solve for $x$ .

Subtract $x^{2}$ from both sides of the equation.

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

Use the quadratic formula to find the solutions.

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

Substitute the values $a = - 1$ , $b = 1$ , and $c = 4$ into the quadratic formula and solve for $x$ .

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

Simplify.

Simplify the numerator.

One to any power is one.

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

Multiply $- 4$ by $- 1$ .

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

Multiply $4$ by $4$ .

$x = \frac{- 1 \pm \sqrt{1 + 16}}{2 \cdot - 1}$

Add $1$ and $16$ .

$x = \frac{- 1 \pm \sqrt{17}}{2 \cdot - 1}$

Multiply $2$ by $- 1$ .

$x = \frac{- 1 \pm \sqrt{17}}{- 2}$

Simplify $\frac{- 1 \pm \sqrt{17}}{- 2}$ .

$x = \frac{1 \pm \sqrt{17}}{2}$

The final answer is the combination of both solutions.

$x = \frac{1 + \sqrt{17}}{2}, \frac{1 - \sqrt{17}}{2}$

Exclude the solutions that do not make $\sqrt{x + 4} = x$ true.

$x = \frac{1 + \sqrt{17}}{2}$

The result can be shown in multiple forms.

Exact Form:

$x = \frac{1 + \sqrt{17}}{2}$

Decimal Form:

$x = 2.56155281 \dots$

Solve for x square root of x+4=x

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