Evaluate square root of x+7-x=1

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

Add $x$ to both sides of the equation.

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

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

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

Simplify each side of the equation.

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

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

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, ${(a^{m})}^{n} = a^{m n}$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

${(x + 7)}^{1} = {(1 + x)}^{2}$

Simplify.

$x + 7 = {(1 + x)}^{2}$

Simplify the right side.

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

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

$x + 7 = (1 + x) (1 + x)$

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

Apply the distributive property.

$x + 7 = 1 (1 + x) + x (1 + x)$

Apply the distributive property.

$x + 7 = 1 \cdot 1 + 1 x + x (1 + x)$

Apply the distributive property.

$x + 7 = 1 \cdot 1 + 1 x + x \cdot 1 + x \cdot x$

Simplify and combine like terms.

Simplify each term.

Multiply $1$ by $1$ .

$x + 7 = 1 + 1 x + x \cdot 1 + x \cdot x$

Multiply $x$ by $1$ .

$x + 7 = 1 + x + x \cdot 1 + x \cdot x$

Multiply $x$ by $1$ .

$x + 7 = 1 + x + x + x \cdot x$

Multiply $x$ by $x$ .

$x + 7 = 1 + x + x + x^{2}$

Add $x$ and $x$ .

$x + 7 = 1 + 2 x + x^{2}$

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.

$1 + 2 x + x^{2} = x + 7$

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

Subtract $x$ from both sides of the equation.

$1 + 2 x + x^{2} - x = 7$

Subtract $x$ from $2 x$ .

$1 + x^{2} + x = 7$

Subtract $7$ from both sides of the equation.

$1 + x^{2} + x - 7 = 0$

Subtract $7$ from $1$ .

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

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 $1$ .

$- 2, 3$

Write the factored form using these integers.

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

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

$x - 2 = 0$

$x + 3 = 0$

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 $x + 3$ equal to $0$ and solve for $x$ .

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

The final solution is all the values that make $(x - 2) (x + 3) = 0$ true.

$x = 2, - 3$

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

$x = 2$

Evaluate square root of x+7-x=1

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