Solve for x x=2 square root of 2x-4

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

Step 1

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

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

Step 2

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

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

Step 3

Simplify each side of the equation.

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

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

Simplify the left side.

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

Apply the product rule to $2 {(2 x - 4)}^{\frac{1}{2}}$ .

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

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

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Apply the distributive property.

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

Multiply.

Multiply $2$ by $4$ .

$8 x + 4 \cdot - 4 = x^{2}$

Multiply $4$ by $- 4$ .

$8 x - 16 = x^{2}$

Step 4

Solve for $x$ .

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

$8 x - 16 - x^{2} = 0$

Factor the left side of the equation.

Factor $- 1$ out of $8 x - 16 - x^{2}$ .

Reorder the expression.

Move $- 16$ .

$8 x - x^{2} - 16 = 0$

Reorder $8 x$ and $- x^{2}$ .

$- x^{2} + 8 x - 16 = 0$

Factor $- 1$ out of $- x^{2}$ .

$- (x^{2}) + 8 x - 16 = 0$

Factor $- 1$ out of $8 x$ .

$- (x^{2}) - (- 8 x) - 16 = 0$

Rewrite $- 16$ as $- 1 (16)$ .

$- (x^{2}) - (- 8 x) - 1 \cdot 16 = 0$

Factor $- 1$ out of $- (x^{2}) - (- 8 x)$ .

$- (x^{2} - 8 x) - 1 \cdot 16 = 0$

Factor $- 1$ out of $- (x^{2} - 8 x) - 1 (16)$ .

$- (x^{2} - 8 x + 16) = 0$

Factor using the perfect square rule.

Rewrite $16$ as $4^{2}$ .

$- (x^{2} - 8 x + 4^{2}) = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$8 x = 2 \cdot x \cdot 4$

Rewrite the polynomial.

$- (x^{2} - 2 \cdot x \cdot 4 + 4^{2}) = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 4$ .

$- {(x - 4)}^{2} = 0$

Divide each term in $- {(x - 4)}^{2} = 0$ by $- 1$ and simplify.

Divide each term in $- {(x - 4)}^{2} = 0$ by $- 1$ .

$\frac{- {(x - 4)}^{2}}{- 1} = \frac{0}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{{(x - 4)}^{2}}{1} = \frac{0}{- 1}$

Divide ${(x - 4)}^{2}$ by $1$ .

${(x - 4)}^{2} = \frac{0}{- 1}$

Simplify the right side.

Divide $0$ by $- 1$ .

${(x - 4)}^{2} = 0$

Set the $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Solve for x x=2 square root of 2x-4

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