Solve for x square root of x+7+2 = square root of 3-x

$\sqrt{x + 7} + 2 = \sqrt{3 - x}$

Step 1

Subtract $2$ from both sides of the equation.

$\sqrt{x + 7} = \sqrt{3 - x} - 2$

Step 2

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

${\sqrt{x + 7}}^{2} = {(\sqrt{3 - x} - 2)}^{2}$

Step 3

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x + 7}$ as ${(x + 7)}^{\frac{1}{2}}$ .

${({(x + 7)}^{\frac{1}{2}})}^{2} = {(\sqrt{3 - x} - 2)}^{2}$

Simplify the left side.

Simplify ${({(x + 7)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(x + 7)}^{\frac{1}{2}})}^{2}$ .

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

${(x + 7)}^{\frac{1}{2} \cdot 2} = {(\sqrt{3 - x} - 2)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(x + 7)}^{\frac{1}{2} \cdot 2} = {(\sqrt{3 - x} - 2)}^{2}$

Rewrite the expression.

${(x + 7)}^{1} = {(\sqrt{3 - x} - 2)}^{2}$

Simplify.

$x + 7 = {(\sqrt{3 - x} - 2)}^{2}$

Simplify the right side.

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

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

$x + 7 = (\sqrt{3 - x} - 2) (\sqrt{3 - x} - 2)$

Expand $(\sqrt{3 - x} - 2) (\sqrt{3 - x} - 2)$ using the FOIL Method.

Apply the distributive property.

$x + 7 = \sqrt{3 - x} (\sqrt{3 - x} - 2) - 2 (\sqrt{3 - x} - 2)$

Apply the distributive property.

$x + 7 = \sqrt{3 - x} \sqrt{3 - x} + \sqrt{3 - x} \cdot - 2 - 2 (\sqrt{3 - x} - 2)$

Apply the distributive property.

$x + 7 = \sqrt{3 - x} \sqrt{3 - x} + \sqrt{3 - x} \cdot - 2 - 2 \sqrt{3 - x} - 2 \cdot - 2$

Simplify and combine like terms.

Simplify each term.

Multiply $\sqrt{3 - x} \sqrt{3 - x}$ .

Raise $\sqrt{3 - x}$ to the power of $1$ .

$x + 7 = {\sqrt{3 - x}}^{1} \sqrt{3 - x} + \sqrt{3 - x} \cdot - 2 - 2 \sqrt{3 - x} - 2 \cdot - 2$

Raise $\sqrt{3 - x}$ to the power of $1$ .

$x + 7 = {\sqrt{3 - x}}^{1} {\sqrt{3 - x}}^{1} + \sqrt{3 - x} \cdot - 2 - 2 \sqrt{3 - x} - 2 \cdot - 2$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$x + 7 = {\sqrt{3 - x}}^{1 + 1} + \sqrt{3 - x} \cdot - 2 - 2 \sqrt{3 - x} - 2 \cdot - 2$

Add $1$ and $1$ .

$x + 7 = {\sqrt{3 - x}}^{2} + \sqrt{3 - x} \cdot - 2 - 2 \sqrt{3 - x} - 2 \cdot - 2$

Rewrite ${\sqrt{3 - x}}^{2}$ as $3 - x$ .

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

$x + 7 = {({(3 - x)}^{\frac{1}{2}})}^{2} + \sqrt{3 - x} \cdot - 2 - 2 \sqrt{3 - x} - 2 \cdot - 2$

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

$x + 7 = {(3 - x)}^{\frac{1}{2} \cdot 2} + \sqrt{3 - x} \cdot - 2 - 2 \sqrt{3 - x} - 2 \cdot - 2$

Combine $\frac{1}{2}$ and $2$ .

$x + 7 = {(3 - x)}^{\frac{2}{2}} + \sqrt{3 - x} \cdot - 2 - 2 \sqrt{3 - x} - 2 \cdot - 2$

Cancel the common factor of $2$ .

Cancel the common factor.

$x + 7 = {(3 - x)}^{\frac{2}{2}} + \sqrt{3 - x} \cdot - 2 - 2 \sqrt{3 - x} - 2 \cdot - 2$

Rewrite the expression.

$x + 7 = {(3 - x)}^{1} + \sqrt{3 - x} \cdot - 2 - 2 \sqrt{3 - x} - 2 \cdot - 2$

Simplify.

$x + 7 = 3 - x + \sqrt{3 - x} \cdot - 2 - 2 \sqrt{3 - x} - 2 \cdot - 2$

Move $- 2$ to the left of $\sqrt{3 - x}$ .

$x + 7 = 3 - x - 2 \cdot \sqrt{3 - x} - 2 \sqrt{3 - x} - 2 \cdot - 2$

Multiply $- 2$ by $- 2$ .

$x + 7 = 3 - x - 2 \sqrt{3 - x} - 2 \sqrt{3 - x} + 4$

Add $3$ and $4$ .

$x + 7 = 7 - x - 2 \sqrt{3 - x} - 2 \sqrt{3 - x}$

Subtract $2 \sqrt{3 - x}$ from $- 2 \sqrt{3 - x}$ .

$x + 7 = 7 - x - 4 \sqrt{3 - x}$

Step 4

Solve for $- 4 \sqrt{3 - x}$ .

Rewrite the equation as $7 - x - 4 \sqrt{3 - x} = x + 7$ .

$7 - x - 4 \sqrt{3 - x} = x + 7$

Move all terms not containing $- 4 \sqrt{3 - x}$ to the right side of the equation.

Subtract $7$ from both sides of the equation.

$- x - 4 \sqrt{3 - x} = x + 7 - 7$

Add $x$ to both sides of the equation.

$- 4 \sqrt{3 - x} = x + 7 - 7 + x$

Combine the opposite terms in $x + 7 - 7 + x$ .

Subtract $7$ from $7$ .

$- 4 \sqrt{3 - x} = x + 0 + x$

Add $x$ and $0$ .

$- 4 \sqrt{3 - x} = x + x$

Add $x$ and $x$ .

$- 4 \sqrt{3 - x} = 2 x$

Step 5

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

${(- 4 \sqrt{3 - x})}^{2} = {(2 x)}^{2}$

Step 6

Simplify each side of the equation.

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

${(- 4 {(3 - x)}^{\frac{1}{2}})}^{2} = {(2 x)}^{2}$

Simplify the left side.

Simplify ${(- 4 {(3 - x)}^{\frac{1}{2}})}^{2}$ .

Apply the product rule to $- 4 {(3 - x)}^{\frac{1}{2}}$ .

${(- 4)}^{2} {({(3 - x)}^{\frac{1}{2}})}^{2} = {(2 x)}^{2}$

Raise $- 4$ to the power of $2$ .

$16 {({(3 - x)}^{\frac{1}{2}})}^{2} = {(2 x)}^{2}$

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

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

$16 {(3 - x)}^{\frac{1}{2} \cdot 2} = {(2 x)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$16 {(3 - x)}^{\frac{1}{2} \cdot 2} = {(2 x)}^{2}$

Rewrite the expression.

$16 {(3 - x)}^{1} = {(2 x)}^{2}$

Simplify.

$16 (3 - x) = {(2 x)}^{2}$

Apply the distributive property.

$16 \cdot 3 + 16 (- x) = {(2 x)}^{2}$

Multiply.

Multiply $16$ by $3$ .

$48 + 16 (- x) = {(2 x)}^{2}$

Multiply $- 1$ by $16$ .

$48 - 16 x = {(2 x)}^{2}$

Simplify the right side.

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

Apply the product rule to $2 x$ .

$48 - 16 x = 2^{2} x^{2}$

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

$48 - 16 x = 4 x^{2}$

Step 7

Solve for $x$ .

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

$48 - 16 x - 4 x^{2} = 0$

Factor the left side of the equation.

Factor $- 4$ out of $48 - 16 x - 4 x^{2}$ .

Reorder the expression.

Move $48$ .

$- 16 x - 4 x^{2} + 48 = 0$

Reorder $- 16 x$ and $- 4 x^{2}$ .

$- 4 x^{2} - 16 x + 48 = 0$

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

$- 4 x^{2} - 16 x + 48 = 0$

Factor $- 4$ out of $- 16 x$ .

$- 4 x^{2} - 4 (4 x) + 48 = 0$

Factor $- 4$ out of $48$ .

$- 4 x^{2} - 4 (4 x) - 4 \cdot - 12 = 0$

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

$- 4 (x^{2} + 4 x) - 4 \cdot - 12 = 0$

Factor $- 4$ out of $- 4 (x^{2} + 4 x) - 4 (- 12)$ .

$- 4 (x^{2} + 4 x - 12) = 0$

Factor.

Factor $x^{2} + 4 x - 12$ 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 $- 12$ and whose sum is $4$ .

$- 2, 6$

Write the factored form using these integers.

$- 4 ((x - 2) (x + 6)) = 0$

Remove unnecessary parentheses.

$- 4 (x - 2) (x + 6) = 0$

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

$x - 2 = 0$

$x + 6 = 0$

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

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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 $- 4 (x - 2) (x + 6) = 0$ true.

$x = 2, - 6$

Step 8

Exclude the solutions that do not make $\sqrt{x + 7} + 2 = \sqrt{3 - x}$ true.

$x = - 6$

Solve for x square root of x+7+2 = square root of 3-x

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