Find the Intersection f(x)=4x^2+2 , f(x)=4x+1

$f (x) = 4 x^{2} + 2$ , $f (x) = 4 x + 1$

Substitute $4 x + 1$ for $f (x)$ .

$4 x + 1 = 4 x^{2} + 2$

Solve for $x$ .

Subtract $4 x^{2}$ from both sides of the equation.

$4 x + 1 - 4 x^{2} = 2$

Subtract $2$ from both sides of the equation.

$4 x + 1 - 4 x^{2} - 2 = 0$

Subtract $2$ from $1$ .

$4 x - 4 x^{2} - 1 = 0$

Factor the left side of the equation.

Factor $- 1$ out of $4 x - 4 x^{2} - 1$ .

Reorder $4 x$ and $- 4 x^{2}$ .

$- 4 x^{2} + 4 x - 1 = 0$

Factor $- 1$ out of $- 4 x^{2}$ .

$- (4 x^{2}) + 4 x - 1 = 0$

Factor $- 1$ out of $4 x$ .

$- (4 x^{2}) - (- 4 x) - 1 = 0$

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

$- (4 x^{2}) - (- 4 x) - 1 \cdot 1 = 0$

Factor $- 1$ out of $- (4 x^{2}) - (- 4 x)$ .

$- (4 x^{2} - 4 x) - 1 \cdot 1 = 0$

Factor $- 1$ out of $- (4 x^{2} - 4 x) - 1 (1)$ .

$- (4 x^{2} - 4 x + 1) = 0$

Factor using the perfect square rule.

Rewrite $4 x^{2}$ as ${(2 x)}^{2}$ .

$- ({(2 x)}^{2} - 4 x + 1) = 0$

Rewrite $1$ as $1^{2}$ .

$- ({(2 x)}^{2} - 4 x + 1^{2}) = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$4 x = 2 \cdot (2 x) \cdot 1$

Rewrite the polynomial.

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

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = 2 x$ and $b = 1$ .

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

Divide each term in $- {(2 x - 1)}^{2} = 0$ by $- 1$ and simplify.

Divide each term in $- {(2 x - 1)}^{2} = 0$ by $- 1$ .

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide ${(2 x - 1)}^{2}$ by $1$ .

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

Simplify the right side.

Divide $0$ by $- 1$ .

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

Set the $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Solve for $x$ .

Add $1$ to both sides of the equation.

$2 x = 1$

Divide each term in $2 x = 1$ by $2$ and simplify.

Divide each term in $2 x = 1$ by $2$ .

$\frac{2 x}{2} = \frac{1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{1}{2}$

Divide $x$ by $1$ .

$x = \frac{1}{2}$

Find the Intersection f(x)=4x^2+2 , f(x)=4x+1

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