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)

Question

Responses

one real solution and three non-real solutions
one real solution and three non-real solutions

two real solutions and two non-real solutions
two real solutions and two non-real solutions

four real solutions
four real solutions

four non-real solutions

Answer 1

To find the solutions of the equation \( 4x^4 - 20x^3 - 81x^2 + 125x + 350 = 0 \), which factors as \( (4x^2 - 25)(x^2 - 5x - 14) = 0 \), we can solve each factor separately.

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

   • This can be rewritten as \( 4x^2 = 25 \).
   • Dividing by 4 gives \( x^2 = 6.25 \), leading to \( x = \pm \sqrt{6.25} = \pm 2.5 \).
   • Thus, there are two real solutions from this factor: \( x = 2.5 \) and \( x = -2.5 \).

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

   • We can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
   • Here, \( a = 1, b = -5, c = -14 \). So, \( b^2 - 4ac = (-5)^2 - 4(1)(-14) = 25 + 56 = 81 \).
   • Since \( \sqrt{81} = 9 \), we find: \[ x = \frac{5 \pm 9}{2} = \frac{14}{2} = 7 \quad \text{and} \quad x = \frac{-4}{2} = -2. \]
   • Hence, this factor also gives two real solutions: \( x = 7 \) and \( x = -2 \).

Combining the solutions from both factors, we have a total of four real solutions: \( 2.5, -2.5, 7, -2 \).

Therefore, the correct response is: four real solutions.