Question

section 3.5 homework

1. solve the system by graphing.

$y = \frac{1}{2}x^2 - 3$
$y = -4 - 2x^2$

a. $\square \left( -\frac{1}{2}, 2 \
ight)$
b. $\square \left( \frac{3}{2}, \frac{4}{3} \
ight)$
c. $\square (7, 2)$
d. $\square (1, 0)$
e. $\square (8, -2)$
f. $\square (0, -4)$
g. $\square \left( \frac{3}{7}, \frac{8}{7} \
ight)$
h. $\square$ no real solution

Explanation:

Step1: Set equations equal

To solve the system \( y = \frac{1}{2}x^2 - 3 \) and \( y = -4 - 2x^2 \), set them equal:
\( \frac{1}{2}x^2 - 3 = -4 - 2x^2 \)

Step2: Solve for \( x^2 \)

Multiply all terms by 2 to eliminate fraction:
\( x^2 - 6 = -8 - 4x^2 \)
Add \( 4x^2 \) and 6 to both sides:
\( 5x^2 = -2 \)

Step3: Analyze solution

The equation \( 5x^2 = -2 \) implies \( x^2 = -\frac{2}{5} \). Since the square of a real number cannot be negative, there are no real solutions for \( x \), so the system has no real solution.

Answer:

F. no real solution