To solve the equation [tex]\(x^4 - 17x^2 + 16 = 0\)[/tex], let's use a substitution method to simplify the problem.
1. Introduce a substitution:
Let [tex]\(u = x^2\)[/tex].
2. Rewrite the original equation:
Substitute [tex]\(u\)[/tex] in place of [tex]\(x^2\)[/tex] in the equation:
[tex]\[
x^4 - 17x^2 + 16 = 0
\][/tex]
becomes,
[tex]\[
u^2 - 17u + 16 = 0
\][/tex]
3. Solve the quadratic equation:
We need to solve the quadratic equation for [tex]\(u\)[/tex]:
[tex]\[
u^2 - 17u + 16 = 0
\][/tex]
To solve this quadratic equation, factorize it:
[tex]\[
u^2 - 17u + 16 = (u - 1)(u - 16) = 0
\][/tex]
4. Find the solutions for [tex]\(u\)[/tex]:
Set each factor equal to zero to find the solutions for [tex]\(u\)[/tex]:
[tex]\[
u - 1 = 0 \quad \Rightarrow \quad u = 1
\][/tex]
[tex]\[
u - 16 = 0 \quad \Rightarrow \quad u = 16
\][/tex]
5. Back-substitute to find [tex]\(x\)[/tex]:
Recall that [tex]\(u = x^2\)[/tex], so we need to solve [tex]\(x^2 = u\)[/tex] for each value of [tex]\(u\)[/tex].
For [tex]\(u = 1\)[/tex]:
[tex]\[
x^2 = 1 \quad \Rightarrow \quad x = \pm 1
\][/tex]
For [tex]\(u = 16\)[/tex]:
[tex]\[
x^2 = 16 \quad \Rightarrow \quad x = \pm 4
\][/tex]
6. Compile the solutions:
The solutions for [tex]\(x\)[/tex] from each value of [tex]\(u\)[/tex] are:
[tex]\[
x = 1, -1, 4, -4
\][/tex]
Thus, the solutions to the equation [tex]\(x^4 - 17x^2 + 16 = 0\)[/tex] are [tex]\(x = 1, -1, 4, -4\)[/tex].
