Question

Which of the following correctly describes the solutions for the equation 4x4−20x3−81x2+125x+350=0 , which can also be written as (4x2−25)(x2−5x−14)=0 ? Do as little calculation as possible.(1 point) Responses four non-real solutions four non-real solutions two real solutions and two non-real solutions two real solutions and two non-real solutions one real solution and three non-real solutions one real solution and three non-real solutions four real solutions

Answer 1

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

1. Factor \( 4x^2 - 25 = 0 \): \[ 4x^2 = 25 \implies x^2 = \frac{25}{4} \implies x = \pm \frac{5}{2} \] This gives us two real solutions: \( x = \frac{5}{2} \) and \( x = -\frac{5}{2} \).

2. Factor \( x^2 - 5x - 14 = 0 \): We can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ a = 1, b = -5, c = -14 \implies b^2 - 4ac = (-5)^2 - 4(1)(-14) = 25 + 56 = 81 \] Since the discriminant (81) is positive, this factor has two real solutions: \[ x = \frac{5 \pm 9}{2} \implies x = 7 \quad \text{and} \quad x = -2 \]

Combining the solutions from both factors, we have the following four real solutions:

1. \( \frac{5}{2} \)
2. \( -\frac{5}{2} \)
3. \( 7 \)
4. \( -2 \)

Thus, the correct description of the solutions for the equation \( 4x^4 - 20x^3 - 81x^2 + 125x + 350 = 0 \) is four real solutions.