Solve for x square root of x+6+1 = square root of 7-x

$\sqrt{x + 6} + 1 = \sqrt{7 - x}$

Step 1

Subtract $1$ from both sides of the equation.

$\sqrt{x + 6} = \sqrt{7 - x} - 1$

Step 2

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

${\sqrt{x + 6}}^{2} = {(\sqrt{7 - x} - 1)}^{2}$

Step 3

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x + 6}$ as ${(x + 6)}^{\frac{1}{2}}$ .

${({(x + 6)}^{\frac{1}{2}})}^{2} = {(\sqrt{7 - x} - 1)}^{2}$

Simplify the left side.

Simplify ${({(x + 6)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(x + 6)}^{\frac{1}{2}})}^{2}$ .

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

${(x + 6)}^{\frac{1}{2} \cdot 2} = {(\sqrt{7 - x} - 1)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(x + 6)}^{\frac{1}{2} \cdot 2} = {(\sqrt{7 - x} - 1)}^{2}$

Rewrite the expression.

${(x + 6)}^{1} = {(\sqrt{7 - x} - 1)}^{2}$

Simplify.

$x + 6 = {(\sqrt{7 - x} - 1)}^{2}$

Simplify the right side.

Simplify ${(\sqrt{7 - x} - 1)}^{2}$ .

Rewrite ${(\sqrt{7 - x} - 1)}^{2}$ as $(\sqrt{7 - x} - 1) (\sqrt{7 - x} - 1)$ .

$x + 6 = (\sqrt{7 - x} - 1) (\sqrt{7 - x} - 1)$

Expand $(\sqrt{7 - x} - 1) (\sqrt{7 - x} - 1)$ using the FOIL Method.

Apply the distributive property.

$x + 6 = \sqrt{7 - x} (\sqrt{7 - x} - 1) - 1 (\sqrt{7 - x} - 1)$

Apply the distributive property.

$x + 6 = \sqrt{7 - x} \sqrt{7 - x} + \sqrt{7 - x} \cdot - 1 - 1 (\sqrt{7 - x} - 1)$

Apply the distributive property.

$x + 6 = \sqrt{7 - x} \sqrt{7 - x} + \sqrt{7 - x} \cdot - 1 - 1 \sqrt{7 - x} - 1 \cdot - 1$

Simplify and combine like terms.

Simplify each term.

Multiply $\sqrt{7 - x} \sqrt{7 - x}$ .

Raise $\sqrt{7 - x}$ to the power of $1$ .

$x + 6 = {\sqrt{7 - x}}^{1} \sqrt{7 - x} + \sqrt{7 - x} \cdot - 1 - 1 \sqrt{7 - x} - 1 \cdot - 1$

Raise $\sqrt{7 - x}$ to the power of $1$ .

$x + 6 = {\sqrt{7 - x}}^{1} {\sqrt{7 - x}}^{1} + \sqrt{7 - x} \cdot - 1 - 1 \sqrt{7 - x} - 1 \cdot - 1$

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

$x + 6 = {\sqrt{7 - x}}^{1 + 1} + \sqrt{7 - x} \cdot - 1 - 1 \sqrt{7 - x} - 1 \cdot - 1$

Add $1$ and $1$ .

$x + 6 = {\sqrt{7 - x}}^{2} + \sqrt{7 - x} \cdot - 1 - 1 \sqrt{7 - x} - 1 \cdot - 1$

Rewrite ${\sqrt{7 - x}}^{2}$ as $7 - x$ .

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

$x + 6 = {({(7 - x)}^{\frac{1}{2}})}^{2} + \sqrt{7 - x} \cdot - 1 - 1 \sqrt{7 - x} - 1 \cdot - 1$

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

$x + 6 = {(7 - x)}^{\frac{1}{2} \cdot 2} + \sqrt{7 - x} \cdot - 1 - 1 \sqrt{7 - x} - 1 \cdot - 1$

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

$x + 6 = {(7 - x)}^{\frac{2}{2}} + \sqrt{7 - x} \cdot - 1 - 1 \sqrt{7 - x} - 1 \cdot - 1$

Cancel the common factor of $2$ .

Cancel the common factor.

$x + 6 = {(7 - x)}^{\frac{2}{2}} + \sqrt{7 - x} \cdot - 1 - 1 \sqrt{7 - x} - 1 \cdot - 1$

Rewrite the expression.

$x + 6 = {(7 - x)}^{1} + \sqrt{7 - x} \cdot - 1 - 1 \sqrt{7 - x} - 1 \cdot - 1$

Simplify.

$x + 6 = 7 - x + \sqrt{7 - x} \cdot - 1 - 1 \sqrt{7 - x} - 1 \cdot - 1$

Move $- 1$ to the left of $\sqrt{7 - x}$ .

$x + 6 = 7 - x - 1 \cdot \sqrt{7 - x} - 1 \sqrt{7 - x} - 1 \cdot - 1$

Rewrite $- 1 \sqrt{7 - x}$ as $- \sqrt{7 - x}$ .

$x + 6 = 7 - x - \sqrt{7 - x} - 1 \sqrt{7 - x} - 1 \cdot - 1$

Rewrite $- 1 \sqrt{7 - x}$ as $- \sqrt{7 - x}$ .

$x + 6 = 7 - x - \sqrt{7 - x} - \sqrt{7 - x} - 1 \cdot - 1$

Multiply $- 1$ by $- 1$ .

$x + 6 = 7 - x - \sqrt{7 - x} - \sqrt{7 - x} + 1$

Add $7$ and $1$ .

$x + 6 = 8 - x - \sqrt{7 - x} - \sqrt{7 - x}$

Subtract $\sqrt{7 - x}$ from $- \sqrt{7 - x}$ .

$x + 6 = 8 - x - 2 \sqrt{7 - x}$

Step 4

Solve for $- 2 \sqrt{7 - x}$ .

Rewrite the equation as $8 - x - 2 \sqrt{7 - x} = x + 6$ .

$8 - x - 2 \sqrt{7 - x} = x + 6$

Move all terms not containing $- 2 \sqrt{7 - x}$ to the right side of the equation.

Subtract $8$ from both sides of the equation.

$- x - 2 \sqrt{7 - x} = x + 6 - 8$

Add $x$ to both sides of the equation.

$- 2 \sqrt{7 - x} = x + 6 - 8 + x$

Add $x$ and $x$ .

$- 2 \sqrt{7 - x} = 2 x + 6 - 8$

Subtract $8$ from $6$ .

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

Step 5

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

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

Step 6

Simplify each side of the equation.

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

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

Simplify the left side.

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

Apply the product rule to $- 2 {(7 - x)}^{\frac{1}{2}}$ .

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

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

$4 {({(7 - x)}^{\frac{1}{2}})}^{2} = {(2 x - 2)}^{2}$

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

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

$4 {(7 - x)}^{\frac{1}{2} \cdot 2} = {(2 x - 2)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$4 {(7 - x)}^{\frac{1}{2} \cdot 2} = {(2 x - 2)}^{2}$

Rewrite the expression.

$4 {(7 - x)}^{1} = {(2 x - 2)}^{2}$

Simplify.

$4 (7 - x) = {(2 x - 2)}^{2}$

Apply the distributive property.

$4 \cdot 7 + 4 (- x) = {(2 x - 2)}^{2}$

Multiply.

Multiply $4$ by $7$ .

$28 + 4 (- x) = {(2 x - 2)}^{2}$

Multiply $- 1$ by $4$ .

$28 - 4 x = {(2 x - 2)}^{2}$

Simplify the right side.

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

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

$28 - 4 x = (2 x - 2) (2 x - 2)$

Expand $(2 x - 2) (2 x - 2)$ using the FOIL Method.

Apply the distributive property.

$28 - 4 x = 2 x (2 x - 2) - 2 (2 x - 2)$

Apply the distributive property.

$28 - 4 x = 2 x (2 x) + 2 x \cdot - 2 - 2 (2 x - 2)$

Apply the distributive property.

$28 - 4 x = 2 x (2 x) + 2 x \cdot - 2 - 2 (2 x) - 2 \cdot - 2$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$28 - 4 x = 2 \cdot 2 x \cdot x + 2 x \cdot - 2 - 2 (2 x) - 2 \cdot - 2$

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

Move $x$ .

$28 - 4 x = 2 \cdot 2 (x \cdot x) + 2 x \cdot - 2 - 2 (2 x) - 2 \cdot - 2$

Multiply $x$ by $x$ .

$28 - 4 x = 2 \cdot 2 x^{2} + 2 x \cdot - 2 - 2 (2 x) - 2 \cdot - 2$

Multiply $2$ by $2$ .

$28 - 4 x = 4 x^{2} + 2 x \cdot - 2 - 2 (2 x) - 2 \cdot - 2$

Multiply $- 2$ by $2$ .

$28 - 4 x = 4 x^{2} - 4 x - 2 (2 x) - 2 \cdot - 2$

Multiply $2$ by $- 2$ .

$28 - 4 x = 4 x^{2} - 4 x - 4 x - 2 \cdot - 2$

Multiply $- 2$ by $- 2$ .

$28 - 4 x = 4 x^{2} - 4 x - 4 x + 4$

Subtract $4 x$ from $- 4 x$ .

$28 - 4 x = 4 x^{2} - 8 x + 4$

Step 7

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.

$4 x^{2} - 8 x + 4 = 28 - 4 x$

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

Add $4 x$ to both sides of the equation.

$4 x^{2} - 8 x + 4 + 4 x = 28$

Add $- 8 x$ and $4 x$ .

$4 x^{2} - 4 x + 4 = 28$

Subtract $28$ from both sides of the equation.

$4 x^{2} - 4 x + 4 - 28 = 0$

Subtract $28$ from $4$ .

$4 x^{2} - 4 x - 24 = 0$

Factor the left side of the equation.

Factor $4$ out of $4 x^{2} - 4 x - 24$ .

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

$4 (x^{2}) - 4 x - 24 = 0$

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

$4 (x^{2}) + 4 (- x) - 24 = 0$

Factor $4$ out of $- 24$ .

$4 (x^{2}) + 4 (- x) + 4 (- 6) = 0$

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

$4 (x^{2} - x) + 4 (- 6) = 0$

Factor $4$ out of $4 (x^{2} - x) + 4 (- 6)$ .

$4 (x^{2} - x - 6) = 0$

Factor.

Factor $x^{2} - x - 6$ 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 $- 1$ .

$- 3, 2$

Write the factored form using these integers.

$4 ((x - 3) (x + 2)) = 0$

Remove unnecessary parentheses.

$4 (x - 3) (x + 2) = 0$

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

$x - 3 = 0$

$x + 2 = 0$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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 $4 (x - 3) (x + 2) = 0$ true.

$x = 3, - 2$

Step 8

Exclude the solutions that do not make $\sqrt{x + 6} + 1 = \sqrt{7 - x}$ true.

$x = - 2$

Solve for x square root of x+6+1 = square root of 7-x

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