Question

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

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 \( x^2 + 4x - 1 = 2x + 2 \), we first rearrange the equation to set it to zero:

\[
x^2 + 4x - 1 - 2x - 2 = 0
\]
This simplifies to:

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

Now we can define two functions based on the original equation:

1. \( f(x) = x^2 + 2x - 3 \)
2. \( g(x) = 0 \)

To find the intersection points of these functions, we can solve the equation:

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

Next, we can factor the quadratic:

\[
(x + 3)(x - 1) = 0
\]

Setting each factor to zero gives us the solutions:

\[
x + 3 = 0 \quad \Rightarrow \quad x = -3
\]
\[
x - 1 = 0 \quad \Rightarrow \quad x = 1
\]

Thus, the solutions to the equation \( x^2 + 4x - 1 = 2x + 2 \) are:

\[
\{-3, 1\}
\]

In conclusion, the two functions to solve this equation would be:

- \( f(x) = x^2 + 2x - 3 \)
- \( g(x) = 0 \)

And the solution set is:

\[
\{-3, 1\}
\]