Solve by Factoring 3x = square root of 20-3x

$3 x = \sqrt{20 - 3 x}$

Step 1

Subtract $\sqrt{20 - 3 x}$ from both sides of the equation.

$3 x - \sqrt{20 - 3 x} = 0$

Step 2

Subtract $3 x$ from both sides of the equation.

$- \sqrt{20 - 3 x} = - 3 x$

Step 3

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

${(- \sqrt{20 - 3 x})}^{2} = {(- 3 x)}^{2}$

Step 4

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{20 - 3 x}$ as ${(20 - 3 x)}^{\frac{1}{2}}$ .

${(- {(20 - 3 x)}^{\frac{1}{2}})}^{2} = {(- 3 x)}^{2}$

Simplify the left side.

Simplify ${(- {(20 - 3 x)}^{\frac{1}{2}})}^{2}$ .

Apply the product rule to $- {(20 - 3 x)}^{\frac{1}{2}}$ .

${(- 1)}^{2} {({(20 - 3 x)}^{\frac{1}{2}})}^{2} = {(- 3 x)}^{2}$

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

$1 {({(20 - 3 x)}^{\frac{1}{2}})}^{2} = {(- 3 x)}^{2}$

Multiply ${({(20 - 3 x)}^{\frac{1}{2}})}^{2}$ by $1$ .

${({(20 - 3 x)}^{\frac{1}{2}})}^{2} = {(- 3 x)}^{2}$

Multiply the exponents in ${({(20 - 3 x)}^{\frac{1}{2}})}^{2}$ .

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

${(20 - 3 x)}^{\frac{1}{2} \cdot 2} = {(- 3 x)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(20 - 3 x)}^{\frac{1}{2} \cdot 2} = {(- 3 x)}^{2}$

Rewrite the expression.

${(20 - 3 x)}^{1} = {(- 3 x)}^{2}$

Simplify.

$20 - 3 x = {(- 3 x)}^{2}$

Simplify the right side.

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

Apply the product rule to $- 3 x$ .

$20 - 3 x = {(- 3)}^{2} x^{2}$

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

$20 - 3 x = 9 x^{2}$

Step 5

Solve for $x$ .

Subtract $9 x^{2}$ from both sides of the equation.

$20 - 3 x - 9 x^{2} = 0$

Factor the left side of the equation.

Factor $- 1$ out of $20 - 3 x - 9 x^{2}$ .

Reorder the expression.

Move $20$ .

$- 3 x - 9 x^{2} + 20 = 0$

Reorder $- 3 x$ and $- 9 x^{2}$ .

$- 9 x^{2} - 3 x + 20 = 0$

Factor $- 1$ out of $- 9 x^{2}$ .

$- (9 x^{2}) - 3 x + 20 = 0$

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

$- (9 x^{2}) - (3 x) + 20 = 0$

Rewrite $20$ as $- 1 (- 20)$ .

$- (9 x^{2}) - (3 x) - 1 \cdot - 20 = 0$

Factor $- 1$ out of $- (9 x^{2}) - (3 x)$ .

$- (9 x^{2} + 3 x) - 1 \cdot - 20 = 0$

Factor $- 1$ out of $- (9 x^{2} + 3 x) - 1 (- 20)$ .

$- (9 x^{2} + 3 x - 20) = 0$

Factor.

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 9 \cdot - 20 = - 180$ and whose sum is $b = 3$ .

Factor $3$ out of $3 x$ .

$- (9 x^{2} + 3 (x) - 20) = 0$

Rewrite $3$ as $- 12$ plus $15$

$- (9 x^{2} + (- 12 + 15) x - 20) = 0$

Apply the distributive property.

$- (9 x^{2} - 12 x + 15 x - 20) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$- ((9 x^{2} - 12 x) + 15 x - 20) = 0$

Factor out the greatest common factor (GCF) from each group.

$- (3 x (3 x - 4) + 5 (3 x - 4)) = 0$

Factor the polynomial by factoring out the greatest common factor, $3 x - 4$ .

$- ((3 x - 4) (3 x + 5)) = 0$

Remove unnecessary parentheses.

$- (3 x - 4) (3 x + 5) = 0$

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

$3 x - 4 = 0$

$3 x + 5 = 0$

Set $3 x - 4$ equal to $0$ and solve for $x$ .

Set $3 x - 4$ equal to $0$ .

$3 x - 4 = 0$

Solve $3 x - 4 = 0$ for $x$ .

Add $4$ to both sides of the equation.

$3 x = 4$

Divide each term in $3 x = 4$ by $3$ and simplify.

Divide each term in $3 x = 4$ by $3$ .

$\frac{3 x}{3} = \frac{4}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{4}{3}$

Divide $x$ by $1$ .

$x = \frac{4}{3}$

Set $3 x + 5$ equal to $0$ and solve for $x$ .

Set $3 x + 5$ equal to $0$ .

$3 x + 5 = 0$

Solve $3 x + 5 = 0$ for $x$ .

Subtract $5$ from both sides of the equation.

$3 x = - 5$

Divide each term in $3 x = - 5$ by $3$ and simplify.

Divide each term in $3 x = - 5$ by $3$ .

$\frac{3 x}{3} = \frac{- 5}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 5}{3}$

Divide $x$ by $1$ .

$x = \frac{- 5}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{5}{3}$

The final solution is all the values that make $- (3 x - 4) (3 x + 5) = 0$ true.

$x = \frac{4}{3}, - \frac{5}{3}$

Step 6

Exclude the solutions that do not make $3 x - \sqrt{20 - 3 x} = 0$ true.

$x = \frac{4}{3}$

Solve by Factoring 3x = square root of 20-3x

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