Solve by Factoring 3x = square root of 8x+1

$3 x = \sqrt{8 x + 1}$

Step 1

Subtract $\sqrt{8 x + 1}$ from both sides of the equation.

$3 x - \sqrt{8 x + 1} = 0$

Step 2

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

$- \sqrt{8 x + 1} = - 3 x$

Step 3

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

${(- \sqrt{8 x + 1})}^{2} = {(- 3 x)}^{2}$

Step 4

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{8 x + 1}$ as ${(8 x + 1)}^{\frac{1}{2}}$ .

${(- {(8 x + 1)}^{\frac{1}{2}})}^{2} = {(- 3 x)}^{2}$

Simplify the left side.

Simplify ${(- {(8 x + 1)}^{\frac{1}{2}})}^{2}$ .

Apply the product rule to $- {(8 x + 1)}^{\frac{1}{2}}$ .

${(- 1)}^{2} {({(8 x + 1)}^{\frac{1}{2}})}^{2} = {(- 3 x)}^{2}$

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

$1 {({(8 x + 1)}^{\frac{1}{2}})}^{2} = {(- 3 x)}^{2}$

Multiply ${({(8 x + 1)}^{\frac{1}{2}})}^{2}$ by $1$ .

${({(8 x + 1)}^{\frac{1}{2}})}^{2} = {(- 3 x)}^{2}$

Multiply the exponents in ${({(8 x + 1)}^{\frac{1}{2}})}^{2}$ .

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

${(8 x + 1)}^{\frac{1}{2} \cdot 2} = {(- 3 x)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(8 x + 1)}^{\frac{1}{2} \cdot 2} = {(- 3 x)}^{2}$

Rewrite the expression.

${(8 x + 1)}^{1} = {(- 3 x)}^{2}$

Simplify.

$8 x + 1 = {(- 3 x)}^{2}$

Simplify the right side.

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

Apply the product rule to $- 3 x$ .

$8 x + 1 = {(- 3)}^{2} x^{2}$

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

$8 x + 1 = 9 x^{2}$

Step 5

Solve for $x$ .

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

$8 x + 1 - 9 x^{2} = 0$

Factor the left side of the equation.

Factor $- 1$ out of $8 x + 1 - 9 x^{2}$ .

Reorder the expression.

Move $1$ .

$8 x - 9 x^{2} + 1 = 0$

Reorder $8 x$ and $- 9 x^{2}$ .

$- 9 x^{2} + 8 x + 1 = 0$

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

$- (9 x^{2}) + 8 x + 1 = 0$

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

$- (9 x^{2}) - (- 8 x) + 1 = 0$

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

$- (9 x^{2}) - (- 8 x) - 1 \cdot - 1 = 0$

Factor $- 1$ out of $- (9 x^{2}) - (- 8 x)$ .

$- (9 x^{2} - 8 x) - 1 \cdot - 1 = 0$

Factor $- 1$ out of $- (9 x^{2} - 8 x) - 1 (- 1)$ .

$- (9 x^{2} - 8 x - 1) = 0$

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 $a \cdot c = 9 \cdot - 1 = - 9$ and whose sum is $b = - 8$ .

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

$- (9 x^{2} - 8 x - 1) = 0$

Rewrite $- 8$ as $1$ plus $- 9$

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

Apply the distributive property.

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

Multiply $x$ by $1$ .

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

$- (x (9 x + 1) - (9 x + 1)) = 0$

Factor the polynomial by factoring out the greatest common factor, $9 x + 1$ .

$- ((9 x + 1) (x - 1)) = 0$

Remove unnecessary parentheses.

$- (9 x + 1) (x - 1) = 0$

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

$9 x + 1 = 0$

$x - 1 = 0$

Set $9 x + 1$ equal to $0$ and solve for $x$ .

Set $9 x + 1$ equal to $0$ .

$9 x + 1 = 0$

Solve $9 x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$9 x = - 1$

Divide each term in $9 x = - 1$ by $9$ and simplify.

Divide each term in $9 x = - 1$ by $9$ .

$\frac{9 x}{9} = \frac{- 1}{9}$

Simplify the left side.

Cancel the common factor of $9$ .

Cancel the common factor.

$\frac{9 x}{9} = \frac{- 1}{9}$

Divide $x$ by $1$ .

$x = \frac{- 1}{9}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{9}$

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.

$x = 1$

The final solution is all the values that make $- (9 x + 1) (x - 1) = 0$ true.

$x = - \frac{1}{9}, 1$

Step 6

Exclude the solutions that do not make $3 x - \sqrt{8 x + 1} = 0$ true.

$x = 1$

Solve by Factoring 3x = square root of 8x+1

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