Question

find the real or imaginary solutions of the equation by factoring.
$x^{4}-17x^{2}=-16$
the solutions are \boxed{}. (use a comma to separate answers as needed.)

Explanation:

Step1: Rewrite the equation

First, we rewrite the given equation \(x^{4}-17x^{2}=- 16\) in standard quadratic form (by letting \(y = x^{2}\)). Add 16 to both sides:
\(x^{4}-17x^{2}+16 = 0\)
Let \(y=x^{2}\), then the equation becomes \(y^{2}-17y + 16=0\)

Step2: Factor the quadratic equation

Factor the quadratic equation \(y^{2}-17y + 16 = 0\). We need two numbers that multiply to 16 and add up to - 17. The numbers are - 1 and - 16.
So, \(y^{2}-17y + 16=(y - 1)(y - 16)=0\)

Step3: Solve for y

Set each factor equal to zero:
\(y-1 = 0\) or \(y - 16=0\)
Solving these, we get \(y = 1\) or \(y=16\)

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

• When \(y = 1\), we have \(x^{2}=1\). Taking square roots of both sides, \(x=\pm\sqrt{1}=\pm1\)
• When \(y = 16\), we have \(x^{2}=16\). Taking square roots of both sides, \(x=\pm\sqrt{16}=\pm4\)

Answer:

\(-4, - 1,1,4\)