Question

3.06 solving polynomial equ...

1. find all real solutions of ( x^4 = 8x^2 + 9 ).

let ( u = x^2 ).
enter your next step here

Explanation:

Step1: Substitute \( u = x^2 \) into the equation

Given the equation \( x^4 = 8x^2 + 9 \), since \( u = x^2 \), then \( x^4=(x^2)^2 = u^2 \). Substituting these into the equation, we get:
\( u^2=8u + 9 \)

Step2: Rearrange into standard quadratic form

Subtract \( 8u + 9 \) from both sides to get a quadratic equation in terms of \( u \):
\( u^2-8u - 9=0 \)

Step3: Factor the quadratic equation

We need to find two numbers that multiply to -9 and add to -8. The numbers are -9 and 1. So,
\( (u - 9)(u+ 1)=0 \)

Step4: Solve for \( u \)

Set each factor equal to zero:

• For \( u - 9=0 \), we get \( u = 9 \)
• For \( u + 1=0 \), we get \( u=-1 \)

Step5: Substitute back \( u = x^2 \) and solve for \( x \)

• When \( u = 9 \), \( x^2=9 \), then \( x=\pm\sqrt{9}=\pm 3 \)
• When \( u=-1 \), \( x^2=-1 \). Since the square of a real number cannot be negative, there are no real solutions for this case.

Answer:

\( x = 3 \) or \( x=-3 \)