Find the Roots (Zeros) square root of x+6+x=6

$\sqrt{x + 6} + x = 6$

Step 1

Subtract $x$ from both sides of the equation.

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

Step 2

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

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

Step 3

Simplify each side of the equation.

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

Simplify ${(6 - x)}^{2}$ .

Rewrite ${(6 - x)}^{2}$ as $(6 - x) (6 - x)$ .

$x + 6 = (6 - x) (6 - x)$

Expand $(6 - x) (6 - x)$ using the FOIL Method.

Apply the distributive property.

$x + 6 = 6 (6 - x) - x (6 - x)$

Apply the distributive property.

$x + 6 = 6 \cdot 6 + 6 (- x) - x (6 - x)$

Apply the distributive property.

$x + 6 = 6 \cdot 6 + 6 (- x) - x \cdot 6 - x (- x)$

Simplify and combine like terms.

Simplify each term.

Multiply $6$ by $6$ .

$x + 6 = 36 + 6 (- x) - x \cdot 6 - x (- x)$

Multiply $- 1$ by $6$ .

$x + 6 = 36 - 6 x - x \cdot 6 - x (- x)$

Multiply $6$ by $- 1$ .

$x + 6 = 36 - 6 x - 6 x - x (- x)$

Rewrite using the commutative property of multiplication.

$x + 6 = 36 - 6 x - 6 x - 1 \cdot - 1 x \cdot x$

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

Move $x$ .

$x + 6 = 36 - 6 x - 6 x - 1 \cdot - 1 (x \cdot x)$

Multiply $x$ by $x$ .

$x + 6 = 36 - 6 x - 6 x - 1 \cdot - 1 x^{2}$

Multiply $- 1$ by $- 1$ .

$x + 6 = 36 - 6 x - 6 x + 1 x^{2}$

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

$x + 6 = 36 - 6 x - 6 x + x^{2}$

Subtract $6 x$ from $- 6 x$ .

$x + 6 = 36 - 12 x + x^{2}$

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.

$36 - 12 x + x^{2} = x + 6$

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

Subtract $x$ from both sides of the equation.

$36 - 12 x + x^{2} - x = 6$

Subtract $x$ from $- 12 x$ .

$36 + x^{2} - 13 x = 6$

Subtract $6$ from both sides of the equation.

$36 + x^{2} - 13 x - 6 = 0$

Subtract $6$ from $36$ .

$x^{2} - 13 x + 30 = 0$

Factor $x^{2} - 13 x + 30$ 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 $30$ and whose sum is $- 13$ .

$- 10, - 3$

Write the factored form using these integers.

$(x - 10) (x - 3) = 0$

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

$x - 10 = 0$

$x - 3 = 0$

Set $x - 10$ equal to $0$ and solve for $x$ .

Set $x - 10$ equal to $0$ .

$x - 10 = 0$

Add $10$ to both sides of the equation.

$x = 10$

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.

$x = 3$

The final solution is all the values that make $(x - 10) (x - 3) = 0$ true.

$x = 10, 3$

Step 5

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

$x = 3$

Step 6

Find the Roots (Zeros) square root of x+6+x=6

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