Solve for x square root of 19-3x-1=3x

$\sqrt{19 - 3 x} - 1 = 3 x$

Step 1

Add $1$ to both sides of the equation.

$\sqrt{19 - 3 x} = 3 x + 1$

Step 2

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

${\sqrt{19 - 3 x}}^{2} = {(3 x + 1)}^{2}$

Step 3

Simplify each side of the equation.

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

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

Simplify the left side.

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

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

Rewrite ${(3 x + 1)}^{2}$ as $(3 x + 1) (3 x + 1)$ .

$19 - 3 x = (3 x + 1) (3 x + 1)$

Expand $(3 x + 1) (3 x + 1)$ using the FOIL Method.

Apply the distributive property.

$19 - 3 x = 3 x (3 x + 1) + 1 (3 x + 1)$

Apply the distributive property.

$19 - 3 x = 3 x (3 x) + 3 x \cdot 1 + 1 (3 x + 1)$

Apply the distributive property.

$19 - 3 x = 3 x (3 x) + 3 x \cdot 1 + 1 (3 x) + 1 \cdot 1$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$19 - 3 x = 3 \cdot 3 x \cdot x + 3 x \cdot 1 + 1 (3 x) + 1 \cdot 1$

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

Move $x$ .

$19 - 3 x = 3 \cdot 3 (x \cdot x) + 3 x \cdot 1 + 1 (3 x) + 1 \cdot 1$

Multiply $x$ by $x$ .

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

Multiply $3$ by $3$ .

$19 - 3 x = 9 x^{2} + 3 x \cdot 1 + 1 (3 x) + 1 \cdot 1$

Multiply $3$ by $1$ .

$19 - 3 x = 9 x^{2} + 3 x + 1 (3 x) + 1 \cdot 1$

Multiply $3 x$ by $1$ .

$19 - 3 x = 9 x^{2} + 3 x + 3 x + 1 \cdot 1$

Multiply $1$ by $1$ .

$19 - 3 x = 9 x^{2} + 3 x + 3 x + 1$

Add $3 x$ and $3 x$ .

$19 - 3 x = 9 x^{2} + 6 x + 1$

Step 4

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.

$9 x^{2} + 6 x + 1 = 19 - 3 x$

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

Add $3 x$ to both sides of the equation.

$9 x^{2} + 6 x + 1 + 3 x = 19$

Add $6 x$ and $3 x$ .

$9 x^{2} + 9 x + 1 = 19$

Subtract $19$ from both sides of the equation.

$9 x^{2} + 9 x + 1 - 19 = 0$

Subtract $19$ from $1$ .

$9 x^{2} + 9 x - 18 = 0$

Factor the left side of the equation.

Factor $9$ out of $9 x^{2} + 9 x - 18$ .

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

$9 (x^{2}) + 9 x - 18 = 0$

Factor $9$ out of $9 x$ .

$9 (x^{2}) + 9 (x) - 18 = 0$

Factor $9$ out of $- 18$ .

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

Factor $9$ out of $9 (x^{2}) + 9 x$ .

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

Factor $9$ out of $9 (x^{2} + x) + 9 \cdot - 2$ .

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

Factor.

Factor $x^{2} + x - 2$ 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 $- 2$ and whose sum is $1$ .

$- 1, 2$

Write the factored form using these integers.

$9 ((x - 1) (x + 2)) = 0$

Remove unnecessary parentheses.

$9 (x - 1) (x + 2) = 0$

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

$x - 1 = 0$

$x + 2 = 0$

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.

$x = 1$

Set $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

The final solution is all the values that make $9 (x - 1) (x + 2) = 0$ true.

$x = 1, - 2$

Step 5

Exclude the solutions that do not make $\sqrt{19 - 3 x} - 1 = 3 x$ true.

$x = 1$

Solve for x square root of 19-3x-1=3x

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