Question

Solve the equation by making an appropriate substitution.
x Superscript 4 Baseline minus 17 x squared plus 16 equals 0
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Part 1
Make an appropriate substitution and rewrite the equation in quadratic form.
Let uequals
  
enter your response here​, then the quadratic equation in u is
  
enter your response here.
Part 2
Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice.
A.
The solution set is StartSet nothing EndSet .
  
​(Type an integer or a simplified fraction. Use a comma to separate answers as​ needed.)
B.
The solution set is empty set.

Answer 1

To solve the equation \( x^4 - 17x^2 + 16 = 0 \), we can make a substitution. We can let \( u = x^2 \). Then, the equation becomes:

\[
u^2 - 17u + 16 = 0
\]

This is a quadratic equation in \( u \).

### Part 1
So we have:
- Let \( u = x^2 \).

Then the quadratic equation in \( u \) is:

\[
u^2 - 17u + 16 = 0
\]

### Part 2
To solve the quadratic equation, we can apply the quadratic formula:

\[
u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

where \( a = 1, b = -17, c = 16 \).

Calculating the discriminant:

\[
b^2 - 4ac = (-17)^2 - 4(1)(16) = 289 - 64 = 225
\]

Now we can find \( u \):

\[
u = \frac{17 \pm \sqrt{225}}{2 \cdot 1} = \frac{17 \pm 15}{2}
\]

Calculating both possibilities:

1. \( u = \frac{17 + 15}{2} = \frac{32}{2} = 16 \)
2. \( u = \frac{17 - 15}{2} = \frac{2}{2} = 1 \)

Now we revert back to \( x^2 \):

1. If \( u = 16 \), then \( x^2 = 16 \) → \( x = 4 \) or \( x = -4 \).
2. If \( u = 1 \), then \( x^2 = 1 \) → \( x = 1 \) or \( x = -1 \).

Thus, the solutions for \( x \) are:

\[
x = 4, x = -4, x = 1, x = -1
\]

### Solution Set
The solution set is:

\[
\{ -4, -1, 1, 4 \}
\]

### Selecting the Correct Choice
This means the answer will be:

**A.** The solution set is \(\{-4, -1, 1, 4\}\).