Solve by Factoring 2x = square root of 19x+5

$2 x = \sqrt{19 x + 5}$

Step 1

Subtract $\sqrt{19 x + 5}$ from both sides of the equation.

$2 x - \sqrt{19 x + 5} = 0$

Step 2

Subtract $2 x$ from both sides of the equation.

$- \sqrt{19 x + 5} = - 2 x$

Step 3

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

${(- \sqrt{19 x + 5})}^{2} = {(- 2 x)}^{2}$

Step 4

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{19 x + 5}$ as ${(19 x + 5)}^{\frac{1}{2}}$ .

${(- {(19 x + 5)}^{\frac{1}{2}})}^{2} = {(- 2 x)}^{2}$

Simplify the left side.

Simplify ${(- {(19 x + 5)}^{\frac{1}{2}})}^{2}$ .

Apply the product rule to $- {(19 x + 5)}^{\frac{1}{2}}$ .

${(- 1)}^{2} {({(19 x + 5)}^{\frac{1}{2}})}^{2} = {(- 2 x)}^{2}$

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

$1 {({(19 x + 5)}^{\frac{1}{2}})}^{2} = {(- 2 x)}^{2}$

Multiply ${({(19 x + 5)}^{\frac{1}{2}})}^{2}$ by $1$ .

${({(19 x + 5)}^{\frac{1}{2}})}^{2} = {(- 2 x)}^{2}$

Multiply the exponents in ${({(19 x + 5)}^{\frac{1}{2}})}^{2}$ .

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

${(19 x + 5)}^{\frac{1}{2} \cdot 2} = {(- 2 x)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(19 x + 5)}^{\frac{1}{2} \cdot 2} = {(- 2 x)}^{2}$

Rewrite the expression.

${(19 x + 5)}^{1} = {(- 2 x)}^{2}$

Simplify.

$19 x + 5 = {(- 2 x)}^{2}$

Simplify the right side.

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

Apply the product rule to $- 2 x$ .

$19 x + 5 = {(- 2)}^{2} x^{2}$

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

$19 x + 5 = 4 x^{2}$

Step 5

Solve for $x$ .

Subtract $4 x^{2}$ from both sides of the equation.

$19 x + 5 - 4 x^{2} = 0$

Factor the left side of the equation.

Factor $- 1$ out of $19 x + 5 - 4 x^{2}$ .

Reorder the expression.

Move $5$ .

$19 x - 4 x^{2} + 5 = 0$

Reorder $19 x$ and $- 4 x^{2}$ .

$- 4 x^{2} + 19 x + 5 = 0$

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

$- (4 x^{2}) + 19 x + 5 = 0$

Factor $- 1$ out of $19 x$ .

$- (4 x^{2}) - (- 19 x) + 5 = 0$

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

$- (4 x^{2}) - (- 19 x) - 1 \cdot - 5 = 0$

Factor $- 1$ out of $- (4 x^{2}) - (- 19 x)$ .

$- (4 x^{2} - 19 x) - 1 \cdot - 5 = 0$

Factor $- 1$ out of $- (4 x^{2} - 19 x) - 1 (- 5)$ .

$- (4 x^{2} - 19 x - 5) = 0$

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 $a \cdot c = 4 \cdot - 5 = - 20$ and whose sum is $b = - 19$ .

Factor $- 19$ out of $- 19 x$ .

$- (4 x^{2} - 19 x - 5) = 0$

Rewrite $- 19$ as $1$ plus $- 20$

$- (4 x^{2} + (1 - 20) x - 5) = 0$

Apply the distributive property.

$- (4 x^{2} + 1 x - 20 x - 5) = 0$

Multiply $x$ by $1$ .

$- (4 x^{2} + x - 20 x - 5) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$- ((4 x^{2} + x) - 20 x - 5) = 0$

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

$- (x (4 x + 1) - 5 (4 x + 1)) = 0$

Factor the polynomial by factoring out the greatest common factor, $4 x + 1$ .

$- ((4 x + 1) (x - 5)) = 0$

Remove unnecessary parentheses.

$- (4 x + 1) (x - 5) = 0$

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

$4 x + 1 = 0$

$x - 5 = 0$

Set $4 x + 1$ equal to $0$ and solve for $x$ .

Set $4 x + 1$ equal to $0$ .

$4 x + 1 = 0$

Solve $4 x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$4 x = - 1$

Divide each term in $4 x = - 1$ by $4$ and simplify.

Divide each term in $4 x = - 1$ by $4$ .

$\frac{4 x}{4} = \frac{- 1}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{- 1}{4}$

Divide $x$ by $1$ .

$x = \frac{- 1}{4}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{4}$

Set $x - 5$ equal to $0$ and solve for $x$ .

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

The final solution is all the values that make $- (4 x + 1) (x - 5) = 0$ true.

$x = - \frac{1}{4}, 5$

Step 6

Exclude the solutions that do not make $2 x - \sqrt{19 x + 5} = 0$ true.

$x = 5$

Solve by Factoring 2x = square root of 19x+5

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