Question

find the solution set for each open sentence.

1. $y^2 - 36 = 0$
2. $x^3 + 6x = 7x^2$

Explanation:

Step 1: Solve $y^2 - 36 = 0$

Rearrange to isolate $y^2$:
$y^2 = 36$
Take square roots of both sides:
$y = \pm\sqrt{36} = \pm6$

Step 2: Rearrange $x^3 + 6x = 7x^2$

Bring all terms to one side:
$x^3 - 7x^2 + 6x = 0$

Step 3: Factor the cubic equation

Factor out $x$ first:
$x(x^2 - 7x + 6) = 0$
Factor the quadratic:
$x(x - 1)(x - 6) = 0$

Step 4: Find roots of cubic equation

Set each factor equal to 0:
$x = 0$, $x - 1 = 0 \implies x=1$, $x - 6 = 0 \implies x=6$

Answer:

For problem 44: $\{-6, 6\}$
For problem 45: $\{0, 1, 6\}$