Question

1. solve for x: ( x^4 - 20x^2 + 64 = 0 ). let p be equal to ( x^2 ).

Explanation:

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

The equation \( x^4 - 20x^2 + 64 = 0 \) can be rewritten using \( p = x^2 \) (since \( x^4=(x^2)^2 = p^2 \)). So the equation becomes \( p^2 - 20p + 64 = 0 \).

Step2: Factor the quadratic in \( p \)

We need to find two numbers that multiply to \( 64 \) and add up to \( -20 \). The numbers are \( -16 \) and \( -4 \). So we can factor the quadratic as \( (p - 16)(p - 4)=0 \).

Step3: Solve for \( p \)

Set each factor equal to zero:

• \( p - 16 = 0 \) gives \( p = 16 \)
• \( p - 4 = 0 \) gives \( p = 4 \)

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

• For \( p = 16 \): \( x^2 = 16 \), taking square roots on both sides, we get \( x=\pm\sqrt{16}=\pm 4 \)
• For \( p = 4 \): \( x^2 = 4 \), taking square roots on both sides, we get \( x=\pm\sqrt{4}=\pm 2 \)

Answer:

The solutions for \( x \) are \( x = 4, x = -4, x = 2, x = -2 \)