Question

Which of the following correctly describes the solutions for the equation 4x^4 - 20x^3 - 81x^2 + 125x + 350 = 0, which can also be written as (4x^2 - 25)(x^2 - 5x - 14) = 0?
Do as little calculation as possible.

A. One real solution and three non-real solutions
B. Two real solutions and two non-real solutions
C. Four non-real solutions
D. Four real solutions

Answer 1

To analyze the equation \( (4x^2 - 25)(x^2 - 5x - 14) = 0 \), we can solve each factor separately.

1. **For the first factor \( 4x^2 - 25 = 0 \):**

Rearranging gives:
\[
4x^2 = 25 \quad \Rightarrow \quad x^2 = \frac{25}{4}
\]
Taking the square root results in:
\[
x = \frac{5}{2} \quad \text{or} \quad x = -\frac{5}{2}
\]
Hence, there are 2 real solutions from this factor.

2. **For the second factor \( x^2 - 5x - 14 = 0 \):**

We can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = -5, \) and \( c = -14 \):
\[
x = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot (-14)}}{2 \cdot 1} = \frac{5 \pm \sqrt{25 + 56}}{2} = \frac{5 \pm \sqrt{81}}{2} = \frac{5 \pm 9}{2}
\]
This results in:
\[
x = \frac{14}{2} = 7 \quad \text{and} \quad x = \frac{-4}{2} = -2
\]
Thus, there are also 2 real solutions from this factor.

In total, combining both factors:
- From \( 4x^2 - 25 = 0 \), we have 2 real solutions.
- From \( x^2 - 5x - 14 = 0 \), we have another 2 real solutions.

Thus, there are a total of 4 real solutions to the original equation.

The correct answer is **D. Four real solutions.**