Evaluate square root of 3-2x=x

$\sqrt{3 - 2 x} = x$

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

${\sqrt{3 - 2 x}}^{2} = x^{2}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3 - 2 x}$ as ${(3 - 2 x)}^{\frac{1}{2}}$ .

${({(3 - 2 x)}^{\frac{1}{2}})}^{2} = x^{2}$

Simplify the left side.

Simplify ${({(3 - 2 x)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(3 - 2 x)}^{\frac{1}{2}})}^{2}$ .

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

${(3 - 2 x)}^{\frac{1}{2} \cdot 2} = x^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(3 - 2 x)}^{\frac{1}{2} \cdot 2} = x^{2}$

Rewrite the expression.

${(3 - 2 x)}^{1} = x^{2}$

Simplify.

$3 - 2 x = x^{2}$

Solve for $x$ .

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

$3 - 2 x - x^{2} = 0$

Factor the left side of the equation.

Factor $- 1$ out of $3 - 2 x - x^{2}$ .

Reorder the expression.

Move $3$ .

$- 2 x - x^{2} + 3 = 0$

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

$- x^{2} - 2 x + 3 = 0$

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

$- (x^{2}) - 2 x + 3 = 0$

Factor $- 1$ out of $- 2 x$ .

$- (x^{2}) - (2 x) + 3 = 0$

Rewrite $3$ as $- 1 (- 3)$ .

$- (x^{2}) - (2 x) - 1 \cdot - 3 = 0$

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

$- (x^{2} + 2 x) - 1 \cdot - 3 = 0$

Factor $- 1$ out of $- (x^{2} + 2 x) - 1 (- 3)$ .

$- (x^{2} + 2 x - 3) = 0$

Factor.

Factor $x^{2} + 2 x - 3$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 3$ and whose sum is $2$ .

$- 1, 3$

Write the factored form using these integers.

$- ((x - 1) (x + 3)) = 0$

Remove unnecessary parentheses.

$- (x - 1) (x + 3) = 0$

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

$x - 1 = 0$

$x + 3 = 0$

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

The final solution is all the values that make $- (x - 1) (x + 3) = 0$ true.

$x = 1, - 3$

Exclude the solutions that do not make $\sqrt{3 - 2 x} = x$ true.

$x = 1$

Evaluate square root of 3-2x=x

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