Let's work through the problems in the image.

1. Resolving Equations in Quadratic Form:

a) $2x^4 + 17x^2 - 9 = 0$

1. Let $y = x^2$, then rewrite the equation: $2y^2 + 17y - 9 = 0$

2. Use the quadratic formula to solve for $y$: $y = \frac{-17 \pm \sqrt{17^2 - 4(2)(-9)}}{2(2)} = \frac{-17 \pm \sqrt{289 + 72}}{4} = \frac{-17 \pm \sqrt{361}}{4} = \frac{-17 \pm 19}{4}$ So, $y = \frac{-17 + 19}{4} = \frac{2}{4} = \frac{1}{2}$ or $y = \frac{-17 - 19}{4} = \frac{-36}{4} = -9$

3. Recall that $y = x^2$, so $x^2 = \frac{1}{2}$ or $x^2 = -9$.

4. Since $x^2 = -9$ has no real solution, the real solution comes from $x^2 = \frac{1}{2}$, which gives: $x = \pm \sqrt{\frac{1}{2}} = \pm \frac{\sqrt{2}}{2}$

Thus, the real solutions are $x = \pm \frac{\sqrt{2}}{2}$.

b) $\frac{1}{x^2} - \frac{7}{x} - 1 = 0$

Let $y = \frac{1}{x}$, then the equation becomes: $y^2 - 7y - 1 = 0$ Use the quadratic formula: $y = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(1)(-1)}}{2(1)} = \frac{7 \pm \sqrt{49 + 4}}{2} = \frac{7 \pm \sqrt{53}}{2}$ So, $y = \frac{7 \pm \sqrt{53}}{2}$.

Since $y = \frac{1}{x}$, we have: $x = \frac{2}{7 \pm \sqrt{53}}$

c) $2w^4 + 2w^2 - 1 = 0$

Let $z = w^2$, then the equation becomes: $2z^2 + 2z - 1 = 0$ Use the quadratic formula: $z = \frac{-2 \pm \sqrt{2^2 - 4(2)(-1)}}{2(2)} = \frac{-2 \pm \sqrt{4 + 8}}{4} = \frac{-2 \pm \sqrt{12}}{4} = \frac{-2 \pm 2\sqrt{3}}{4} = \frac{-1 \pm \sqrt{3}}{2}$ So, $z = \frac{-1 + \sqrt{3}}{2}$ or $z = \frac{-1 - \sqrt{3}}{2}$.

Since $z = w^2$, we need the real solutions, which only come from $z = \frac{-1 + \sqrt{3}}{2}$.

Thus, $w^2 = \frac{-1 + \sqrt{3}}{2}$, so $w = \pm \sqrt{\frac{-1 + \sqrt{3}}{2}}$.

d) $\sqrt{x+2} - \sqrt{x-1} = 1$

Isolate one of the square roots: $\sqrt{x+2} = \sqrt{x-1} + 1$ Square both sides: $x+2 = (\sqrt{x-1} + 1)^2 = (x-1) + 2\sqrt{x-1} + 1 = x + 2\sqrt{x-1}$ This simplifies to: $2 = 2\sqrt{x-1}$ $1 = \sqrt{x-1}$ Square both sides again: $1 = x-1$ So, $x = 2$.

2. Solve the following equations:

a) $\sqrt{x-1} + \sqrt{x-4} = 5$

Isolate one square root: $\sqrt{x-1} = 5 - \sqrt{x-4}$ Square both sides: $x - 1 = (5 - \sqrt{x-4})^2 = 25 - 10\sqrt{x-4} + (x - 4)$ This simplifies to: $x - 1 = x + 21 - 10\sqrt{x-4}$ Simplifying further: $-22 = -10\sqrt{x-4}$ $\sqrt{x-4} = \frac{22}{10} = 2.2$ Square both sides again: $x - 4 = 2.2^2 = 4.84$ Thus, $x = 8.84$.

Would you like more detailed solutions, or do you have specific questions about these problems? Here are five questions for practice:

1. How can you factor a quadratic equation?
2. What are the possible cases when solving a quadratic equation?
3. What is the significance of the discriminant in the quadratic formula?
4. How do you isolate square roots when solving radical equations?
5. Why is it important to check for extraneous solutions when solving equations involving square roots?

Tip: Always check your solutions in the original equation to ensure there are no extraneous results.