Question

find the real solutions of the following equation.
$x^4 - 20x^2 + 64 = 0$

select the correct choice below and, if necessary, fill in the answer box to complete your choice.

a. the solution set is { }.
(use a comma to separate answers as needed.)

b. there are no real solutions.

Explanation:

Step1: Substitute \( y = x^2 \)

Let \( y = x^2 \), then the equation \( x^4 - 20x^2 + 64 = 0 \) becomes \( y^2 - 20y + 64 = 0 \).

Step2: Factor the quadratic equation

Factor \( y^2 - 20y + 64 \). We need two numbers that multiply to 64 and add to -20. Those numbers are -16 and -4. So, \( y^2 - 20y + 64=(y - 16)(y - 4)=0 \).

Step3: Solve for \( y \)

Set each factor equal to zero: \( y - 16 = 0 \) or \( y - 4 = 0 \). So, \( y = 16 \) or \( y = 4 \).

Step4: Substitute back \( y = x^2 \) and solve for \( x \)

• For \( y = 16 \): \( x^2 = 16 \), so \( x=\pm\sqrt{16}=\pm 4 \).
• For \( y = 4 \): \( x^2 = 4 \), so \( x=\pm\sqrt{4}=\pm 2 \).

Answer:

-4, -2, 2, 4