Given the two quadratic equations:

I. $x^2 - 14x + 24 = 0$

II. $y^2 + 8y + 15 = 0$

What is the highest root of equation I as a percentage of the lowest root of equation II?

Question

Given the two quadratic equations:

I. $x^2 - 14x + 24 = 0$

II. $y^2 + 8y + 15 = 0$

What is the highest root of equation I as a percentage of the lowest root of equation II?

Accepted Answer

**Solution:**

1. Solve equation I: $x^2 - 14x + 24 = 0$

- Calculate the discriminant:
     $$D = (-14)^2 - 4 \times 1 \times 24 = 196 - 96 = 100$$
   - Roots are:
     $$x = \frac{14 \pm \sqrt{100}}{2} = \frac{14 \pm 10}{2}$$
   - So, roots are:
     $$x_1 = \frac{14 + 10}{2} = 12, \quad x_2 = \frac{14 - 10}{2} = 2$$
   - Highest root of equation I is $12$.

2. Solve equation II: $y^2 + 8y + 15 = 0$

- Calculate the discriminant:
     $$D = 8^2 - 4 \times 1 \times 15 = 64 - 60 = 4$$
   - Roots are:
     $$y = \frac{-8 \pm \sqrt{4}}{2} = \frac{-8 \pm 2}{2}$$
   - So, roots are:
     $$y_1 = \frac{-8 + 2}{2} = -3, \quad y_2 = \frac{-8 - 2}{2} = -5$$
   - Lowest root of equation II is $-5$.

3. Find the highest root of equation I as a percentage of the lowest root of equation II:

$$
   \text{Percentage} = \left( \frac{12}{-5} \right) \times 100 = -240\% 
   $$

4. However, the question asks for the highest root of equation I as a percentage of the lowest root of equation II. Since the lowest root is negative, the percentage is negative.

5. Among the options, $-240\%$ is given, but the correct answer should be $-240\%$.

**Answer:** $-240\%$

**Note:** The original options include $125\%$ as correct, but based on calculations, the answer is $-240\%$. Please verify the question context or options if the sign is considered differently.

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