Find All Complex Solutions square root of 1-3x=x+3

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

Step 1

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

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

Step 2

Simplify each side of the equation.

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

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

Simplify the left side.

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

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

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

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

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

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

$1 - 3 x = x \cdot x + x \cdot 3 + 3 x + 3 \cdot 3$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

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

Move $3$ to the left of $x$ .

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

Multiply $3$ by $3$ .

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

Add $3 x$ and $3 x$ .

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

Step 3

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.

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

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

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

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

Add $6 x$ and $3 x$ .

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

Subtract $1$ from both sides of the equation.

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

Subtract $1$ from $9$ .

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

Factor $x^{2} + 9 x + 8$ 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 $8$ and whose sum is $9$ .

$1, 8$

Write the factored form using these integers.

$(x + 1) (x + 8) = 0$

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

$x + 1 = 0$

$x + 8 = 0$

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 $x + 8$ equal to $0$ and solve for $x$ .

Set $x + 8$ equal to $0$ .

$x + 8 = 0$

Subtract $8$ from both sides of the equation.

$x = - 8$

The final solution is all the values that make $(x + 1) (x + 8) = 0$ true.

$x = - 1, - 8$

Step 4

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

$x = - 1$

Find All Complex Solutions square root of 1-3x=x+3

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