The image shows a mathematical task involving solving three quadratic equations. Let's solve them step by step:

i) $x^2 + 5x - 14 = 0$

To solve this quadratic equation, we use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here:

• $a = 1$,
• $b = 5$,
• $c = -14$.

Substitute these values into the formula: $x = \frac{-5 \pm \sqrt{5^2 - 4(1)(-14)}}{2(1)}$ $x = \frac{-5 \pm \sqrt{25 + 56}}{2}$ $x = \frac{-5 \pm \sqrt{81}}{2}$ $x = \frac{-5 \pm 9}{2}$

Now, calculate the two solutions:

1. $x_1 = \frac{-5 + 9}{2} = \frac{4}{2} = 2$,
2. $x_2 = \frac{-5 - 9}{2} = \frac{-14}{2} = -7$.

Solution for i: $x = 2$ or $x = -7$.

ii) $x^2 + 8x + 16 = 0$

This is a perfect square trinomial: $x^2 + 8x + 16 = (x + 4)^2 = 0$ Thus: $x + 4 = 0 \implies x = -4$

Solution for ii: $x = -4$ (repeated root).

iii) $-4x^2 - 5x = 0$

First, factor out $-x$: $-4x^2 - 5x = 0 \implies -x(4x + 5) = 0$ This gives two solutions:

1. $-x = 0 \implies x = 0$,
2. $4x + 5 = 0 \implies x = -\frac{5}{4}$.

Solution for iii: $x = 0$ or $x = -\frac{5}{4}$.

Final Answers:

i) $x = 2$ or $x = -7$
ii) $x = -4$
iii) $x = 0$ or $x = -\frac{5}{4}$

Let me know if you'd like further clarification or details!

Follow-up Questions:

1. What is the quadratic formula, and why is it used?
2. How can you recognize a perfect square trinomial?
3. How do you factorize quadratic equations efficiently?
4. What is the significance of a "repeated root"?
5. When should you factor out terms in an equation?

Tip:

Always check for simple factoring options first before using the quadratic formula—it can save time!