Solve by Factoring square root of x+4+2=x

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

Step 1

Subtract $x$ from both sides of the equation.

$\sqrt{x + 4} + 2 - x = 0$

Step 2

Move all terms not containing $\sqrt{x + 4}$ to the right side of the equation.

Subtract $2$ from both sides of the equation.

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

Add $x$ to both sides of the equation.

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

Step 3

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

${\sqrt{x + 4}}^{2} = {(- 2 + x)}^{2}$

Step 4

Simplify each side of the equation.

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

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

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, ${(a^{m})}^{n} = a^{m n}$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

Simplify ${(- 2 + x)}^{2}$ .

Rewrite ${(- 2 + x)}^{2}$ as $(- 2 + x) (- 2 + x)$ .

$x + 4 = (- 2 + x) (- 2 + x)$

Expand $(- 2 + x) (- 2 + x)$ using the FOIL Method.

Apply the distributive property.

$x + 4 = - 2 (- 2 + x) + x (- 2 + x)$

Apply the distributive property.

$x + 4 = - 2 \cdot - 2 - 2 x + x (- 2 + x)$

Apply the distributive property.

$x + 4 = - 2 \cdot - 2 - 2 x + x \cdot - 2 + x \cdot x$

Simplify and combine like terms.

Simplify each term.

Multiply $- 2$ by $- 2$ .

$x + 4 = 4 - 2 x + x \cdot - 2 + x \cdot x$

Move $- 2$ to the left of $x$ .

$x + 4 = 4 - 2 x - 2 \cdot x + x \cdot x$

Multiply $x$ by $x$ .

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

Subtract $2 x$ from $- 2 x$ .

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

Step 5

Solve for $x$ .

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

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

Move all terms containing $x$ to the left side of the equation.

Subtract $x$ from both sides of the equation.

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

Subtract $x$ from $- 4 x$ .

$4 + x^{2} - 5 x = 4$

Subtract $4$ from both sides of the equation.

$4 + x^{2} - 5 x - 4 = 0$

Combine the opposite terms in $4 + x^{2} - 5 x - 4$ .

Subtract $4$ from $4$ .

$x^{2} - 5 x + 0 = 0$

Add $x^{2} - 5 x$ and $0$ .

$x^{2} - 5 x = 0$

Factor $x$ out of $x^{2} - 5 x$ .

Factor $x$ out of $x^{2}$ .

$x \cdot x - 5 x = 0$

Factor $x$ out of $- 5 x$ .

$x \cdot x + x \cdot - 5 = 0$

Factor $x$ out of $x \cdot x + x \cdot - 5$ .

$x (x - 5) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$x - 5 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x - 5$ equal to $0$ and solve for $x$ .

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

The final solution is all the values that make $x (x - 5) = 0$ true.

$x = 0, 5$

Step 6

Exclude the solutions that do not make $\sqrt{x + 4} + 2 - x = 0$ true.

$x = 5$

Solve by Factoring square root of x+4+2=x

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