Solve by Factoring 3x = square root of 5+4x

$3 x = \sqrt{5 + 4 x}$

Move $\sqrt{5 + 4 x}$ to the left side of the equation by subtracting it from both sides.

$3 x - \sqrt{5 + 4 x} = 0$

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

$- \sqrt{5 + 4 x} = - 3 x$

Multiply each term in $- \sqrt{5 + 4 x} = - 3 x$ by $- 1$

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Multiply each term in $- \sqrt{5 + 4 x} = - 3 x$ by $- 1$ .

$(- \sqrt{5 + 4 x}) \cdot - 1 = (- 3 x) \cdot - 1$

Multiply $- \sqrt{5 + 4 x} \cdot - 1$ .

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Multiply $- 1$ by $- 1$ .

$1 \sqrt{5 + 4 x} = (- 3 x) \cdot - 1$

Multiply $\sqrt{5 + 4 x}$ by $1$ .

$\sqrt{5 + 4 x} = (- 3 x) \cdot - 1$

Multiply $- 1$ by $- 3$ .

$\sqrt{5 + 4 x} = 3 x$

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

${\sqrt{5 + 4 x}}^{2} = {(3 x)}^{2}$

Simplify each side of the equation.

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Multiply the exponents in ${({(5 + 4 x)}^{\frac{1}{2}})}^{2}$ .

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

${(5 + 4 x)}^{\frac{1}{2} \cdot 2} = {(3 x)}^{2}$

Cancel the common factor of $2$ .

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Cancel the common factor.

${(5 + 4 x)}^{\frac{1}{2} \cdot 2} = {(3 x)}^{2}$

Rewrite the expression.

${(5 + 4 x)}^{1} = {(3 x)}^{2}$

Simplify.

$5 + 4 x = {(3 x)}^{2}$

Apply the product rule to $3 x$ .

$5 + 4 x = 3^{2} x^{2}$

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

$5 + 4 x = 9 x^{2}$

Solve for $x$ .

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

$5 + 4 x - 9 x^{2} = 0$

Factor the left side of the equation.

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Factor $- 1$ out of $5 + 4 x - 9 x^{2}$ .

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Reorder the expression.

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Move $5$ .

$4 x - 9 x^{2} + 5 = 0$

Reorder $4 x$ and $- 9 x^{2}$ .

$- 9 x^{2} + 4 x + 5 = 0$

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

$- (9 x^{2}) + 4 x + 5 = 0$

Factor $- 1$ out of $4 x$ .

$- (9 x^{2}) - (- 4 x) + 5 = 0$

Rewrite $5$ as $- 1 (- 5)$ .

$- (9 x^{2}) - (- 4 x) - 1 \cdot - 5 = 0$

Factor $- 1$ out of $- (9 x^{2}) - (- 4 x)$ .

$- (9 x^{2} - 4 x) - 1 \cdot - 5 = 0$

Factor $- 1$ out of $- (9 x^{2} - 4 x) - 1 (- 5)$ .

$- (9 x^{2} - 4 x - 5) = 0$

Factor.

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Factor by grouping.

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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 $a \cdot c = 9 \cdot - 5 = - 45$ and whose sum is $b = - 4$ .

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Factor $- 4$ out of $- 4 x$ .

$- (9 x^{2} - 4 x - 5) = 0$

Rewrite $- 4$ as $5$ plus $- 9$

$- (9 x^{2} + (5 - 9) x - 5) = 0$

Apply the distributive property.

$- (9 x^{2} + 5 x - 9 x - 5) = 0$

Factor out the greatest common factor from each group.

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Group the first two terms and the last two terms.

$- ((9 x^{2} + 5 x) - 9 x - 5) = 0$

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

$- (x (9 x + 5) - (9 x + 5)) = 0$

Factor the polynomial by factoring out the greatest common factor, $9 x + 5$ .

$- ((9 x + 5) (x - 1)) = 0$

Remove unnecessary parentheses.

$- (9 x + 5) (x - 1) = 0$

Multiply each term in $- (9 x + 5) (x - 1) = 0$ by $- 1$

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Multiply each term in $- (9 x + 5) (x - 1) = 0$ by $- 1$ .

$(- (9 x + 5) (x - 1)) \cdot - 1 = 0 \cdot - 1$

Simplify $(- (9 x + 5) (x - 1)) \cdot - 1$ .

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Simplify by multiplying through.

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Apply the distributive property.

$(- (9 x) - 1 \cdot 5) (x - 1) \cdot - 1 = 0 \cdot - 1$

Multiply.

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Multiply $9$ by $- 1$ .

$(- 9 x - 1 \cdot 5) (x - 1) \cdot - 1 = 0 \cdot - 1$

Multiply $- 1$ by $5$ .

$(- 9 x - 5) (x - 1) \cdot - 1 = 0 \cdot - 1$

Expand $(- 9 x - 5) (x - 1)$ using the FOIL Method.

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Apply the distributive property.

$(- 9 x (x - 1) - 5 (x - 1)) \cdot - 1 = 0 \cdot - 1$

Apply the distributive property.

$(- 9 x \cdot x - 9 x \cdot - 1 - 5 (x - 1)) \cdot - 1 = 0 \cdot - 1$

Apply the distributive property.

$(- 9 x \cdot x - 9 x \cdot - 1 - 5 x - 5 \cdot - 1) \cdot - 1 = 0 \cdot - 1$

Simplify and combine like terms.

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Simplify each term.

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Multiply $x$ by $x$ by adding the exponents.

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Move $x$ .

$(- 9 (x \cdot x) - 9 x \cdot - 1 - 5 x - 5 \cdot - 1) \cdot - 1 = 0 \cdot - 1$

Multiply $x$ by $x$ .

$(- 9 x^{2} - 9 x \cdot - 1 - 5 x - 5 \cdot - 1) \cdot - 1 = 0 \cdot - 1$

Multiply $- 1$ by $- 9$ .

$(- 9 x^{2} + 9 x - 5 x - 5 \cdot - 1) \cdot - 1 = 0 \cdot - 1$

Multiply $- 5$ by $- 1$ .

$(- 9 x^{2} + 9 x - 5 x + 5) \cdot - 1 = 0 \cdot - 1$

Subtract $5 x$ from $9 x$ .

$(- 9 x^{2} + 4 x + 5) \cdot - 1 = 0 \cdot - 1$

Apply the distributive property.

$- 9 x^{2} \cdot - 1 + 4 x \cdot - 1 + 5 \cdot - 1 = 0 \cdot - 1$

Simplify.

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Multiply $- 1$ by $- 9$ .

$9 x^{2} + 4 x \cdot - 1 + 5 \cdot - 1 = 0 \cdot - 1$

Multiply $- 1$ by $4$ .

$9 x^{2} - 4 x + 5 \cdot - 1 = 0 \cdot - 1$

Multiply $5$ by $- 1$ .

$9 x^{2} - 4 x - 5 = 0 \cdot - 1$

Multiply $0$ by $- 1$ .

$9 x^{2} - 4 x - 5 = 0$

Factor by grouping.

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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 $a \cdot c = 9 \cdot - 5 = - 45$ and whose sum is $b = - 4$ .

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Factor $- 4$ out of $- 4 x$ .

$9 x^{2} - 4 x - 5 = 0$

Rewrite $- 4$ as $5$ plus $- 9$

$9 x^{2} + (5 - 9) x - 5 = 0$

Apply the distributive property.

$9 x^{2} + 5 x - 9 x - 5 = 0$

Factor out the greatest common factor from each group.

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Group the first two terms and the last two terms.

$(9 x^{2} + 5 x) - 9 x - 5 = 0$

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

$x (9 x + 5) - (9 x + 5) = 0$

Factor the polynomial by factoring out the greatest common factor, $9 x + 5$ .

$(9 x + 5) (x - 1) = 0$

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

$9 x + 5 = 0$

$x - 1 = 0$

Set the first factor equal to $0$ and solve.

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Set the first factor equal to $0$ .

$9 x + 5 = 0$

Subtract $5$ from both sides of the equation.

$9 x = - 5$

Divide each term by $9$ and simplify.

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Divide each term in $9 x = - 5$ by $9$ .

$\frac{9 x}{9} = \frac{- 5}{9}$

Cancel the common factor of $9$ .

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Cancel the common factor.

$\frac{9 x}{9} = \frac{- 5}{9}$

Divide $x$ by $1$ .

$x = \frac{- 5}{9}$

Move the negative in front of the fraction.

$x = - \frac{5}{9}$

Set the next factor equal to $0$ and solve.

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Set the next factor 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 $(9 x + 5) (x - 1) = 0$ true.

$x = - \frac{5}{9}, 1$

Exclude the solutions that do not make $3 x - \sqrt{5 + 4 x} = 0$ true.

$x = 1$

Solve by Factoring 3x = square root of 5+4x

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