To show how to solve the equation, create two functions and find their intersection points. What two functions can be used to solve the following equation, and what is their solution set?

Question

x^2+4x-1=2x+2

Answers:
f(x) = x^2 + 4x, 9 (x) = 2x, {-3,1}
f(x) = x^2 4x x2+4x (x) = 2x + 2,{-1,3}
f(x) = x^2 +4x 9(x) = (x) + 2x+2,{-3,1}
f(x) = x^2 + 4x-1,9 (x) = 2x +2, {-3, -4, 1,4}

Answer 1

To solve the equation \( x^2 + 4x - 1 = 2x + 2 \) using the intersection of two functions, we start by rearranging the equation into the form \( f(x) = g(x) \).

1. Rearranging the equation:

   \[ x^2 + 4x - 1 - 2x - 2 = 0 \] \[ x^2 + 2x - 3 = 0 \]

Now, we can define our two functions:

• \( f(x) = x^2 + 4x - 1 \)
• \( g(x) = 2x + 2 \)

Next, we will find the intersection points of these two functions by setting them equal to each other:

\[ x^2 + 4x - 1 = 2x + 2 \]

Rearranging:

\[ x^2 + 4x - 2x - 1 - 2 = 0 \] \[ x^2 + 2x - 3 = 0 \]

This is a quadratic equation that we can solve using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

Here, \( a = 1 \), \( b = 2 \), \( c = -3 \).

Calculating the discriminant:

\[ D = b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot (-3) = 4 + 12 = 16 \]

Now, applying the quadratic formula:

\[ x = \frac{-2 \pm \sqrt{16}}{2 \cdot 1} \] \[ x = \frac{-2 \pm 4}{2} \]

Calculating the two possible solutions:

1. \( x = \frac{-2 + 4}{2} = \frac{2}{2} = 1 \)
2. \( x = \frac{-2 - 4}{2} = \frac{-6}{2} = -3 \)

Thus, the solution set for the intersection points is:

\[ {-3, 1} \]

The correct choice from the given answers is:

f(x) = x^2 + 4x - 1, g(x) = 2x + 2, {-3, 1}.