Solve by Factoring x=3 square root of 2x-9

$x = 3 \sqrt{2 x - 9}$

Step 1

Subtract $3 \sqrt{2 x - 9}$ from both sides of the equation.

$x - 3 \sqrt{2 x - 9} = 0$

Step 2

Subtract $x$ from both sides of the equation.

$- 3 \sqrt{2 x - 9} = - x$

Step 3

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

${(- 3 \sqrt{2 x - 9})}^{2} = {(- x)}^{2}$

Step 4

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2 x - 9}$ as ${(2 x - 9)}^{\frac{1}{2}}$ .

${(- 3 {(2 x - 9)}^{\frac{1}{2}})}^{2} = {(- x)}^{2}$

Simplify the left side.

Simplify ${(- 3 {(2 x - 9)}^{\frac{1}{2}})}^{2}$ .

Apply the product rule to $- 3 {(2 x - 9)}^{\frac{1}{2}}$ .

${(- 3)}^{2} {({(2 x - 9)}^{\frac{1}{2}})}^{2} = {(- x)}^{2}$

Raise $- 3$ to the power of $2$ .

$9 {({(2 x - 9)}^{\frac{1}{2}})}^{2} = {(- x)}^{2}$

Multiply the exponents in ${({(2 x - 9)}^{\frac{1}{2}})}^{2}$ .

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

$9 {(2 x - 9)}^{\frac{1}{2} \cdot 2} = {(- x)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$9 {(2 x - 9)}^{\frac{1}{2} \cdot 2} = {(- x)}^{2}$

Rewrite the expression.

$9 {(2 x - 9)}^{1} = {(- x)}^{2}$

Simplify.

$9 (2 x - 9) = {(- x)}^{2}$

Apply the distributive property.

$9 (2 x) + 9 \cdot - 9 = {(- x)}^{2}$

Multiply.

Multiply $2$ by $9$ .

$18 x + 9 \cdot - 9 = {(- x)}^{2}$

Multiply $9$ by $- 9$ .

$18 x - 81 = {(- x)}^{2}$

Simplify the right side.

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

Apply the product rule to $- x$ .

$18 x - 81 = {(- 1)}^{2} x^{2}$

Raise $- 1$ to the power of $2$ .

$18 x - 81 = 1 x^{2}$

Multiply $x^{2}$ by $1$ .

$18 x - 81 = x^{2}$

Step 5

Solve for $x$ .

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

$18 x - 81 - x^{2} = 0$

Factor the left side of the equation.

Factor $- 1$ out of $18 x - 81 - x^{2}$ .

Reorder the expression.

Move $- 81$ .

$18 x - x^{2} - 81 = 0$

Reorder $18 x$ and $- x^{2}$ .

$- x^{2} + 18 x - 81 = 0$

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

$- (x^{2}) + 18 x - 81 = 0$

Factor $- 1$ out of $18 x$ .

$- (x^{2}) - (- 18 x) - 81 = 0$

Rewrite $- 81$ as $- 1 (81)$ .

$- (x^{2}) - (- 18 x) - 1 \cdot 81 = 0$

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

$- (x^{2} - 18 x) - 1 \cdot 81 = 0$

Factor $- 1$ out of $- (x^{2} - 18 x) - 1 (81)$ .

$- (x^{2} - 18 x + 81) = 0$

Factor using the perfect square rule.

Rewrite $81$ as $9^{2}$ .

$- (x^{2} - 18 x + 9^{2}) = 0$

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

$18 x = 2 \cdot x \cdot 9$

Rewrite the polynomial.

$- (x^{2} - 2 \cdot x \cdot 9 + 9^{2}) = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 9$ .

$- {(x - 9)}^{2} = 0$

Divide each term in $- {(x - 9)}^{2} = 0$ by $- 1$ and simplify.

Divide each term in $- {(x - 9)}^{2} = 0$ by $- 1$ .

$\frac{- {(x - 9)}^{2}}{- 1} = \frac{0}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{{(x - 9)}^{2}}{1} = \frac{0}{- 1}$

Divide ${(x - 9)}^{2}$ by $1$ .

${(x - 9)}^{2} = \frac{0}{- 1}$

Simplify the right side.

Divide $0$ by $- 1$ .

${(x - 9)}^{2} = 0$

Set the $x - 9$ equal to $0$ .

$x - 9 = 0$

Add $9$ to both sides of the equation.

$x = 9$

Solve by Factoring x=3 square root of 2x-9

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