Solve by Factoring square root of 3x+19=x+5

$\sqrt{3 x + 19} = x + 5$

Step 1

Move all the expressions to the left side of the equation.

Subtract $x$ from both sides of the equation.

$\sqrt{3 x + 19} - x = 5$

Subtract $5$ from both sides of the equation.

$\sqrt{3 x + 19} - x - 5 = 0$

Step 2

Move all terms not containing $\sqrt{3 x + 19}$ to the right side of the equation.

Add $x$ to both sides of the equation.

$\sqrt{3 x + 19} - 5 = x$

Add $5$ to both sides of the equation.

$\sqrt{3 x + 19} = x + 5$

Step 3

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

${\sqrt{3 x + 19}}^{2} = {(x + 5)}^{2}$

Step 4

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3 x + 19}$ as ${(3 x + 19)}^{\frac{1}{2}}$ .

${({(3 x + 19)}^{\frac{1}{2}})}^{2} = {(x + 5)}^{2}$

Simplify the left side.

Simplify ${({(3 x + 19)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(3 x + 19)}^{\frac{1}{2}})}^{2}$ .

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

${(3 x + 19)}^{\frac{1}{2} \cdot 2} = {(x + 5)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(3 x + 19)}^{\frac{1}{2} \cdot 2} = {(x + 5)}^{2}$

Rewrite the expression.

${(3 x + 19)}^{1} = {(x + 5)}^{2}$

Simplify.

$3 x + 19 = {(x + 5)}^{2}$

Simplify the right side.

Simplify ${(x + 5)}^{2}$ .

Rewrite ${(x + 5)}^{2}$ as $(x + 5) (x + 5)$ .

$3 x + 19 = (x + 5) (x + 5)$

Expand $(x + 5) (x + 5)$ using the FOIL Method.

Apply the distributive property.

$3 x + 19 = x (x + 5) + 5 (x + 5)$

Apply the distributive property.

$3 x + 19 = x \cdot x + x \cdot 5 + 5 (x + 5)$

Apply the distributive property.

$3 x + 19 = x \cdot x + x \cdot 5 + 5 x + 5 \cdot 5$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$3 x + 19 = x^{2} + x \cdot 5 + 5 x + 5 \cdot 5$

Move $5$ to the left of $x$ .

$3 x + 19 = x^{2} + 5 \cdot x + 5 x + 5 \cdot 5$

Multiply $5$ by $5$ .

$3 x + 19 = x^{2} + 5 x + 5 x + 25$

Add $5 x$ and $5 x$ .

$3 x + 19 = x^{2} + 10 x + 25$

Step 5

Solve for $x$ .

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$x^{2} + 10 x + 25 = 3 x + 19$

Move all terms containing $x$ to the left side of the equation.

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

$x^{2} + 10 x + 25 - 3 x = 19$

Subtract $3 x$ from $10 x$ .

$x^{2} + 7 x + 25 = 19$

Subtract $19$ from both sides of the equation.

$x^{2} + 7 x + 25 - 19 = 0$

Subtract $19$ from $25$ .

$x^{2} + 7 x + 6 = 0$

Factor $x^{2} + 7 x + 6$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $6$ and whose sum is $7$ .

$1, 6$

Write the factored form using these integers.

$(x + 1) (x + 6) = 0$

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

$x + 1 = 0$

$x + 6 = 0$

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Set $x + 6$ equal to $0$ and solve for $x$ .

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

The final solution is all the values that make $(x + 1) (x + 6) = 0$ true.

$x = - 1, - 6$

Step 6

Exclude the solutions that do not make $\sqrt{3 x + 19} - x - 5 = 0$ true.

$x = - 1$

Solve by Factoring square root of 3x+19=x+5

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