Let's verify both questions step by step:

Question 2: $2x^4 - 3x^3 + x^2 = 0$

1. Factor the common term $x^2$: $x^2 (2x^2 - 3x + 1) = 0$

2. Solve $x^2 = 0$: $x = 0$

3. Solve $2x^2 - 3x + 1 = 0$: Use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Substituting $a = 2$, $b = -3$, and $c = 1$: $x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(1)}}{2(2)}$ $x = \frac{3 \pm \sqrt{9 - 8}}{4}$ $x = \frac{3 \pm \sqrt{1}}{4}$ $x = \frac{3 \pm 1}{4}$ Two solutions: $x = \frac{3 + 1}{4} = 1 \quad \text{and} \quad x = \frac{3 - 1}{4} = \frac{1}{2}$

4. Final solutions: $x = 0, \, x = 1, \, x = \frac{1}{2}$

The work in the image has a correct solution for $x = 0$ and $x = \frac{3}{2}$ mistakenly shown. It should be $x = 1$ and $x = \frac{1}{2}$.

Question 4: $x^4 + 4x^2 = 32$

1. Let $y = x^2$: The equation becomes: $y^2 + 4y = 32$

2. Rearrange into standard form: $y^2 + 4y - 32 = 0$

3. Solve using the quadratic formula: $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Substituting $a = 1$, $b = 4$, $c = -32$: $y = \frac{-4 \pm \sqrt{4^2 - 4(1)(-32)}}{2(1)}$ $y = \frac{-4 \pm \sqrt{16 + 128}}{2}$ $y = \frac{-4 \pm \sqrt{144}}{2}$ $y = \frac{-4 \pm 12}{2}$ Two solutions: $y = \frac{-4 + 12}{2} = 4 \quad \text{and} \quad y = \frac{-4 - 12}{2} = -8$

4. Back-substitute $y = x^2$:

   • For $y = 4$: $x^2 = 4 \quad \Rightarrow \quad x = \pm 2$
   • For $y = -8$: No real solutions since $x^2 \geq 0$.

5. Final solutions: $x = 2, \, x = -2$

This solution appears untouched in the image.

Final Analysis:

1. For Question 2, there’s an error in one solution ($x = \frac{3}{2}$).
2. For Question 4, the setup and solution seem correct.

Would you like more details on either question or the steps?
Here are 5 related questions for further exploration:

1. How does factoring simplify solving polynomial equations?
2. What is the quadratic formula, and why does it always work for second-degree equations?
3. How can we check all solutions of $x$ in polynomial equations?
4. What does $x^2 \geq 0$ imply about real solutions?
5. How can substitution simplify solving equations like $x^4 + 4x^2 = 32$?

Tip: Always double-check your solutions by substituting them back into the original equation.