Find All Complex Solutions square root of x+5+x=7

$\sqrt{x + 5} + x = 7$

Step 1

Subtract $x$ from both sides of the equation.

$\sqrt{x + 5} = 7 - x$

Step 2

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

${\sqrt{x + 5}}^{2} = {(7 - x)}^{2}$

Step 3

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x + 5}$ as ${(x + 5)}^{\frac{1}{2}}$ .

${({(x + 5)}^{\frac{1}{2}})}^{2} = {(7 - x)}^{2}$

Simplify the left side.

Simplify ${({(x + 5)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(x + 5)}^{\frac{1}{2}})}^{2}$ .

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

${(x + 5)}^{\frac{1}{2} \cdot 2} = {(7 - x)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(x + 5)}^{\frac{1}{2} \cdot 2} = {(7 - x)}^{2}$

Rewrite the expression.

${(x + 5)}^{1} = {(7 - x)}^{2}$

Simplify.

$x + 5 = {(7 - x)}^{2}$

Simplify the right side.

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

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

$x + 5 = (7 - x) (7 - x)$

Expand $(7 - x) (7 - x)$ using the FOIL Method.

Apply the distributive property.

$x + 5 = 7 (7 - x) - x (7 - x)$

Apply the distributive property.

$x + 5 = 7 \cdot 7 + 7 (- x) - x (7 - x)$

Apply the distributive property.

$x + 5 = 7 \cdot 7 + 7 (- x) - x \cdot 7 - x (- x)$

Simplify and combine like terms.

Simplify each term.

Multiply $7$ by $7$ .

$x + 5 = 49 + 7 (- x) - x \cdot 7 - x (- x)$

Multiply $- 1$ by $7$ .

$x + 5 = 49 - 7 x - x \cdot 7 - x (- x)$

Multiply $7$ by $- 1$ .

$x + 5 = 49 - 7 x - 7 x - x (- x)$

Rewrite using the commutative property of multiplication.

$x + 5 = 49 - 7 x - 7 x - 1 \cdot - 1 x \cdot x$

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

Move $x$ .

$x + 5 = 49 - 7 x - 7 x - 1 \cdot - 1 (x \cdot x)$

Multiply $x$ by $x$ .

$x + 5 = 49 - 7 x - 7 x - 1 \cdot - 1 x^{2}$

Multiply $- 1$ by $- 1$ .

$x + 5 = 49 - 7 x - 7 x + 1 x^{2}$

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

$x + 5 = 49 - 7 x - 7 x + x^{2}$

Subtract $7 x$ from $- 7 x$ .

$x + 5 = 49 - 14 x + x^{2}$

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.

$49 - 14 x + x^{2} = x + 5$

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

Subtract $x$ from both sides of the equation.

$49 - 14 x + x^{2} - x = 5$

Subtract $x$ from $- 14 x$ .

$49 + x^{2} - 15 x = 5$

Subtract $5$ from both sides of the equation.

$49 + x^{2} - 15 x - 5 = 0$

Subtract $5$ from $49$ .

$x^{2} - 15 x + 44 = 0$

Factor $x^{2} - 15 x + 44$ 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 $44$ and whose sum is $- 15$ .

$- 11, - 4$

Write the factored form using these integers.

$(x - 11) (x - 4) = 0$

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

$x - 11 = 0$

$x - 4 = 0$

Set $x - 11$ equal to $0$ and solve for $x$ .

Set $x - 11$ equal to $0$ .

$x - 11 = 0$

Add $11$ to both sides of the equation.

$x = 11$

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

The final solution is all the values that make $(x - 11) (x - 4) = 0$ true.

$x = 11, 4$

Step 5

Exclude the solutions that do not make $\sqrt{x + 5} + x = 7$ true.

$x = 4$

Find All Complex Solutions square root of x+5+x=7

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