To solve the equation [tex]\( x^4 - 17x^2 + 16 = 0 \)[/tex], let's use a substitution method. We'll start by letting [tex]\( u = x^2 \)[/tex].
Substituting [tex]\( u = x^2 \)[/tex] into the original equation, we get:
[tex]\[ x^4 - 17x^2 + 16 = 0 \][/tex]
which can be rewritten in terms of [tex]\( u \)[/tex] as:
[tex]\[ u^2 - 17u + 16 = 0 \][/tex]
Now, we have a quadratic equation in [tex]\( u \)[/tex]:
[tex]\[ u^2 - 17u + 16 = 0 \][/tex]
To solve this quadratic equation, we 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].
First, we calculate the discriminant:
[tex]\[ \text{Discriminant} = b^2 - 4ac = (-17)^2 - 4 \cdot 1 \cdot 16 = 289 - 64 = 225 \][/tex]
Next, we find the two 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]
Thus, the solutions for [tex]\( u \)[/tex] are [tex]\( u = 16 \)[/tex] and [tex]\( u = 1 \)[/tex].
Since [tex]\( u = x^2 \)[/tex], we substitute back to find the solutions for [tex]\( x \)[/tex]:
For [tex]\( u = 16 \)[/tex]:
[tex]\[ x^2 = 16 \][/tex]
[tex]\[ x = \sqrt{16} \][/tex]
[tex]\[ x = 4 \quad \text{or} \quad x = -4 \][/tex]
For [tex]\( u = 1 \)[/tex]:
[tex]\[ x^2 = 1 \][/tex]
[tex]\[ x = \sqrt{1} \][/tex]
[tex]\[ x = 1 \quad \text{or} \quad x = -1 \][/tex]
Therefore, the solutions to the original equation [tex]\( x^4 - 17x^2 + 16 = 0 \)[/tex] are:
[tex]\[ x = 4, -4, 1, -1 \][/tex]
