Find All Complex Solutions 6- square root of x-2=x+4

$6 - \sqrt{x - 2} = x + 4$

Step 1

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

Subtract $6$ from both sides of the equation.

$- \sqrt{x - 2} = x + 4 - 6$

Subtract $6$ from $4$ .

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

Step 2

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

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

Step 3

Simplify each side of the equation.

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

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

Simplify the left side.

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

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

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

Raise $- 1$ to the power of $2$ .

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

Multiply ${({(x - 2)}^{\frac{1}{2}})}^{2}$ by $1$ .

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

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

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

$x - 2 = (x - 2) (x - 2)$

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

Apply the distributive property.

$x - 2 = x (x - 2) - 2 (x - 2)$

Apply the distributive property.

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

Apply the distributive property.

$x - 2 = x \cdot x + x \cdot - 2 - 2 x - 2 \cdot - 2$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$x - 2 = x^{2} + x \cdot - 2 - 2 x - 2 \cdot - 2$

Move $- 2$ to the left of $x$ .

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

Multiply $- 2$ by $- 2$ .

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

Subtract $2 x$ from $- 2 x$ .

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

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.

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

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

Subtract $x$ from both sides of the equation.

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

Subtract $x$ from $- 4 x$ .

$x^{2} - 5 x + 4 = - 2$

Add $2$ to both sides of the equation.

$x^{2} - 5 x + 4 + 2 = 0$

Add $4$ and $2$ .

$x^{2} - 5 x + 6 = 0$

Factor $x^{2} - 5 x + 6$ 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 $- 5$ .

$- 3, - 2$

Write the factored form using these integers.

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

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 $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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

$x = 3, 2$

Step 5

Exclude the solutions that do not make $6 - \sqrt{x - 2} = x + 4$ true.

$x = 2$

Find All Complex Solutions 6- square root of x-2=x+4

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