Find the Intersection of the Functions f(x) = square root of 2x+8 , f(x)=x

$f (x) = \sqrt{2 x + 8}$ , $f (x) = x$

Step 1

Substitute $x$ for $f (x)$ .

$x = \sqrt{2 x + 8}$

Step 2

Solve for $x$ .

Since the radical is on the right side of the equation, switch the sides so it is on the left side of the equation.

$\sqrt{2 x + 8} = x$

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

${\sqrt{2 x + 8}}^{2} = x^{2}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2 x + 8}$ as ${(2 x + 8)}^{\frac{1}{2}}$ .

${({(2 x + 8)}^{\frac{1}{2}})}^{2} = x^{2}$

Simplify the left side.

Simplify ${({(2 x + 8)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(2 x + 8)}^{\frac{1}{2}})}^{2}$ .

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

${(2 x + 8)}^{\frac{1}{2} \cdot 2} = x^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(2 x + 8)}^{\frac{1}{2} \cdot 2} = x^{2}$

Rewrite the expression.

${(2 x + 8)}^{1} = x^{2}$

Simplify.

$2 x + 8 = x^{2}$

Solve for $x$ .

Subtract $x^{2}$ from both sides of the equation.

$2 x + 8 - x^{2} = 0$

Factor the left side of the equation.

Factor $- 1$ out of $2 x + 8 - x^{2}$ .

Reorder the expression.

Move $8$ .

$2 x - x^{2} + 8 = 0$

Reorder $2 x$ and $- x^{2}$ .

$- x^{2} + 2 x + 8 = 0$

Factor $- 1$ out of $- x^{2}$ .

$- (x^{2}) + 2 x + 8 = 0$

Factor $- 1$ out of $2 x$ .

$- (x^{2}) - (- 2 x) + 8 = 0$

Rewrite $8$ as $- 1 (- 8)$ .

$- (x^{2}) - (- 2 x) - 1 \cdot - 8 = 0$

Factor $- 1$ out of $- (x^{2}) - (- 2 x)$ .

$- (x^{2} - 2 x) - 1 \cdot - 8 = 0$

Factor $- 1$ out of $- (x^{2} - 2 x) - 1 (- 8)$ .

$- (x^{2} - 2 x - 8) = 0$

Factor.

Factor $x^{2} - 2 x - 8$ 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 $- 8$ and whose sum is $- 2$ .

$- 4, 2$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

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

$x - 4 = 0$

$x + 2 = 0$

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.

$x = 4$

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

$x = 4, - 2$

Exclude the solutions that do not make $x = \sqrt{2 x + 8}$ true.

$x = 4$

Find the Intersection of the Functions f(x) = square root of 2x+8 , f(x)=x

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