To solve the equation [tex]\( x^4 + 3x^2 - 4 = 0 \)[/tex], let's proceed with the following steps:
1. Substitution:
Let [tex]\( y = x^2 \)[/tex]. This substitution transforms the original quartic equation into a quadratic one:
[tex]\[
y^2 + 3y - 4 = 0.
\][/tex]
2. Solve the quadratic equation:
We can solve the quadratic equation [tex]\( y^2 + 3y - 4 = 0 \)[/tex] using the quadratic formula [tex]\( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]. Here, [tex]\( a = 1 \)[/tex], [tex]\( b = 3 \)[/tex], and [tex]\( c = -4 \)[/tex], so:
[tex]\[
y = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} = \frac{-3 \pm \sqrt{9 + 16}}{2} = \frac{-3 \pm \sqrt{25}}{2}.
\][/tex]
Simplifying further, we get:
[tex]\[
y = \frac{-3 \pm 5}{2}.
\][/tex]
3. Find the roots for [tex]\( y \)[/tex]:
This results in two solutions for [tex]\( y \)[/tex]:
[tex]\[
y_1 = \frac{-3 + 5}{2} = 1
\][/tex]
[tex]\[
y_2 = \frac{-3 - 5}{2} = -4.
\][/tex]
4. Back-substitute [tex]\( y \)[/tex] with [tex]\( x^2 \)[/tex]:
Recall that [tex]\( y = x^2 \)[/tex]. Therefore, we have:
[tex]\[
x^2 = 1 \quad \text{or} \quad x^2 = -4.
\][/tex]
5. Solve for [tex]\( x \)[/tex]:
- For [tex]\( x^2 = 1 \)[/tex]:
[tex]\[
x = \pm 1.
\][/tex]
- For [tex]\( x^2 = -4 \)[/tex]:
[tex]\[
x = \pm 2i.
\][/tex]
(Here, [tex]\( i \)[/tex] is the imaginary unit with [tex]\( i^2 = -1 \)[/tex]).
6. Summarize the solutions:
Combining all the solutions, we get:
[tex]\[
x = \pm 1, \quad x = \pm 2i.
\][/tex]
Thus, the solutions to the equation [tex]\( x^4 + 3x^2 - 4 = 0 \)[/tex] are:
[tex]\[
x = -1, \quad x = 1, \quad x = -2i, \quad x = 2i.
\][/tex]
