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Solve [tex]\( x^4 - 17x^2 + 16 = 0 \)[/tex].

Let [tex]\( u = \square \)[/tex].

Sagot :

To solve the equation [tex]\( x^4 - 17x^2 + 16 = 0 \)[/tex], we can use a substitution to simplify the problem.

1. Substitute:
Let [tex]\( u = x^2 \)[/tex].
This means [tex]\( u^2 = (x^2)^2 = x^4 \)[/tex].

2. Rewrite the equation:
Substitute [tex]\( u \)[/tex] into the original equation:
[tex]\[ x^4 - 17x^2 + 16 = 0 \implies u^2 - 17u + 16 = 0 \][/tex]
Now, we have a quadratic equation in terms of [tex]\( u \)[/tex].

3. Solve the quadratic equation:
To solve [tex]\( u^2 - 17u + 16 = 0 \)[/tex], we can use the quadratic formula [tex]\( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -17 \)[/tex], and [tex]\( c = 16 \)[/tex].
- Calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac = (-17)^2 - 4 \cdot 1 \cdot 16 = 289 - 64 = 225 \][/tex]
- Find the solutions for [tex]\( u \)[/tex]:
[tex]\[ u_1 = \frac{17 + \sqrt{225}}{2 \cdot 1} = \frac{17 + 15}{2} = \frac{32}{2} = 16 \][/tex]
[tex]\[ u_2 = \frac{17 - \sqrt{225}}{2 \cdot 1} = \frac{17 - 15}{2} = \frac{2}{2} = 1 \][/tex]

4. Back-substitute to find [tex]\( x \)[/tex]:
Since [tex]\( u = x^2 \)[/tex], we need to solve for [tex]\( x \)[/tex]:
- For [tex]\( u_1 = 16 \)[/tex]:
[tex]\[ x^2 = 16 \implies x = \pm \sqrt{16} = \pm 4 \][/tex]
- For [tex]\( u_2 = 1 \)[/tex]:
[tex]\[ x^2 = 1 \implies x = \pm \sqrt{1} = \pm 1 \][/tex]

5. List all solutions:
The solutions to the original equation [tex]\( x^4 - 17x^2 + 16 = 0 \)[/tex] are:
[tex]\[ x = 4, -4, 1, -1 \][/tex]

So, the complete set of solutions is:
[tex]\[ x = \{4, -4, 1, -1\} \][/tex]