Question

algebra 2a semester online practice
complete this assessment to review what you’ve learned.
graph the system to determine the solutions.
$y = x^2 - 2$
$y = 2x + 1$
(1 point)
○ no solution.
○ $(-3, 7)$ and $(1, -1)$
○ $(-1.414, 0)$ and $(1.414, 0)$
○ $(-1, -1)$ and $(3, 7)$
check answer remaining attempts 3

Explanation:

Step 1: Set equations equal

To find solutions, set \( y = x^2 - 2 \) and \( y = 2x + 1 \) equal:
\( x^2 - 2 = 2x + 1 \)

Step 2: Rearrange into quadratic

Simplify to standard quadratic form (\( ax^2 + bx + c = 0 \)):
\( x^2 - 2x - 3 = 0 \)

Step 3: Solve quadratic equation

Factor the quadratic:
\( (x - 3)(x + 1) = 0 \)
Thus, \( x = 3 \) or \( x = -1 \).

Step 4: Find corresponding y-values

For \( x = 3 \): \( y = 2(3) + 1 = 7 \) (point: \( (3, 7) \))
For \( x = -1 \): \( y = 2(-1) + 1 = -1 \) (point: \( (-1, -1) \))

Step 5: Verify with options

The solutions \( (-1, -1) \) and \( (3, 7) \) match one of the given options.

Answer:

\( (-1, -1) \) and \( (3, 7) \) (the last option: \( (-1, -1) \) and \( (3, 7) \))