Solve by Factoring square root of 18-3x=x

$\sqrt{18 - 3 x} = x$

Step 1

Subtract $x$ from both sides of the equation.

$\sqrt{18 - 3 x} - x = 0$

Step 2

Factor $3$ out of $18 - 3 x$ .

Factor $3$ out of $18$ .

$\sqrt{3 (6) - 3 x} - x = 0$

Factor $3$ out of $- 3 x$ .

$\sqrt{3 (6) + 3 (- x)} - x = 0$

Factor $3$ out of $3 (6) + 3 (- x)$ .

$\sqrt{3 (6 - x)} - x = 0$

Step 3

Add $x$ to both sides of the equation.

$\sqrt{3 (6 - x)} = x$

Step 4

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

${\sqrt{3 (6 - x)}}^{2} = x^{2}$

Step 5

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3 (6 - x)}$ as ${(3 (6 - x))}^{\frac{1}{2}}$ .

${({(3 (6 - x))}^{\frac{1}{2}})}^{2} = x^{2}$

Simplify the left side.

Simplify ${({(3 (6 - x))}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(3 (6 - x))}^{\frac{1}{2}})}^{2}$ .

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

${(3 (6 - x))}^{\frac{1}{2} \cdot 2} = x^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(3 (6 - x))}^{\frac{1}{2} \cdot 2} = x^{2}$

Rewrite the expression.

${(3 (6 - x))}^{1} = x^{2}$

Apply the distributive property.

${(3 \cdot 6 + 3 (- x))}^{1} = x^{2}$

Multiply.

Multiply $3$ by $6$ .

${(18 + 3 (- x))}^{1} = x^{2}$

Multiply $- 1$ by $3$ .

${(18 - 3 x)}^{1} = x^{2}$

Simplify.

$18 - 3 x = x^{2}$

Step 6

Solve for $x$ .

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

$18 - 3 x - x^{2} = 0$

Factor the left side of the equation.

Factor $- 1$ out of $18 - 3 x - x^{2}$ .

Reorder the expression.

Move $18$ .

$- 3 x - x^{2} + 18 = 0$

Reorder $- 3 x$ and $- x^{2}$ .

$- x^{2} - 3 x + 18 = 0$

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

$- (x^{2}) - 3 x + 18 = 0$

Factor $- 1$ out of $- 3 x$ .

$- (x^{2}) - (3 x) + 18 = 0$

Rewrite $18$ as $- 1 (- 18)$ .

$- (x^{2}) - (3 x) - 1 \cdot - 18 = 0$

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

$- (x^{2} + 3 x) - 1 \cdot - 18 = 0$

Factor $- 1$ out of $- (x^{2} + 3 x) - 1 (- 18)$ .

$- (x^{2} + 3 x - 18) = 0$

Factor.

Factor $x^{2} + 3 x - 18$ 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 $- 18$ and whose sum is $3$ .

$- 3, 6$

Write the factored form using these integers.

$- ((x - 3) (x + 6)) = 0$

Remove unnecessary parentheses.

$- (x - 3) (x + 6) = 0$

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

$x - 3 = 0$

$x + 6 = 0$

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Set $x + 6$ equal to $0$ and solve for $x$ .

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

The final solution is all the values that make $- (x - 3) (x + 6) = 0$ true.

$x = 3, - 6$

Step 7

Exclude the solutions that do not make $\sqrt{3 (6 - x)} - x = 0$ true.

$x = 3$

Solve by Factoring square root of 18-3x=x

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