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

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

Step 1

Subtract $x$ from both sides of the equation.

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

Step 2

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

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

Step 3

Simplify each side of the equation.

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

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

Simplify the left side.

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

Apply the product rule to $- {(3 - x)}^{\frac{1}{2}}$ .

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

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

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

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

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

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

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

$3 - x = (- 3 - x) (- 3 - x)$

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

Apply the distributive property.

$3 - x = - 3 (- 3 - x) - x (- 3 - x)$

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Multiply $- 3$ by $- 3$ .

$3 - x = 9 - 3 (- x) - x \cdot - 3 - x (- x)$

Multiply $- 1$ by $- 3$ .

$3 - x = 9 + 3 x - x \cdot - 3 - x (- x)$

Multiply $- 3$ by $- 1$ .

$3 - x = 9 + 3 x + 3 x - x (- x)$

Rewrite using the commutative property of multiplication.

$3 - x = 9 + 3 x + 3 x - 1 \cdot - 1 x \cdot x$

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

Move $x$ .

$3 - x = 9 + 3 x + 3 x - 1 \cdot - 1 (x \cdot x)$

Multiply $x$ by $x$ .

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

Multiply $- 1$ by $- 1$ .

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

Multiply $x^{2}$ by $1$ .

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

Add $3 x$ and $3 x$ .

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

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.

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

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

Add $x$ to both sides of the equation.

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

Add $6 x$ and $x$ .

$9 + x^{2} + 7 x = 3$

Subtract $3$ from both sides of the equation.

$9 + x^{2} + 7 x - 3 = 0$

Subtract $3$ from $9$ .

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

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 5

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

$x = - 1$

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

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