Question

1. what are the complex zeros?

x⁴ - 5x² - 36 = 0
if necessary, enter your answers separate with a comma. no spaces.

show your work check

1. what are the real zeros?

x⁴ - 5x² - 36 = 0
if necessary, enter your answers separate with a comma. no spaces.

Explanation:

Question 5: Complex Zeros of \( x^4 - 5x^2 - 36 = 0 \)

Step 1: Substitute \( y = x^2 \)

Let \( y = x^2 \), then the equation becomes \( y^2 - 5y - 36 = 0 \).

Step 2: Solve the quadratic equation for \( y \)

Factor the quadratic: \( y^2 - 5y - 36 = (y - 9)(y + 4) = 0 \).
Set each factor to zero:
\( y - 9 = 0 \implies y = 9 \)
\( y + 4 = 0 \implies y = -4 \)

Step 3: Substitute back \( y = x^2 \) and solve for \( x \)

• For \( y = 9 \): \( x^2 = 9 \implies x = \pm 3 \) (real zeros, not complex here).
• For \( y = -4 \): \( x^2 = -4 \implies x = \pm \sqrt{-4} = \pm 2i \) (complex zeros).

Step 1: Substitute \( y = x^2 \)

Let \( y = x^2 \), then the equation becomes \( y^2 - 5y - 36 = 0 \).

Step 2: Solve the quadratic equation for \( y \)

Factor the quadratic: \( y^2 - 5y - 36 = (y - 9)(y + 4) = 0 \).
Set each factor to zero:
\( y - 9 = 0 \implies y = 9 \)
\( y + 4 = 0 \implies y = -4 \)

Step 3: Substitute back \( y = x^2 \) and solve for \( x \)

• For \( y = 9 \): \( x^2 = 9 \implies x = \pm 3 \) (real zeros).
• For \( y = -4 \): \( x^2 = -4 \) has no real solutions (since square of real number cannot be negative).

Answer:

\( 2i,-2i \)

Question 6: Real Zeros of \( x^4 - 5x^2 - 36 = 0 \)