Math Solve equations Substitution

# Substitution

Biquadratic equations are special 4th degree polynomial equations of the form:

$ax^4+bx^2+c=0$

Biquadratic equations can be converted into quadratic equations by substitution $x^2=z$.

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### Hint

The quadratic equation resulting from the substitution can be solved with the quadratic formula.

### Example

Solve the biquadratic equation: $x^4-3x^2+2=0$

1. #### Substitution

The given equation is substituted by replacing $x^2$ with $z$.
$x^4-3x^2+2=0$

$z=x^2$
$z^2-3z+2=0$
2. #### Solve the quadratic equation

The new quadratic equation can now be solved e.g. with the pq-formula.
$z^2-3z+2=0$

$z_{1,2} = \frac{p}{2} \pm\sqrt{(\frac{p}{2})^2-q}$

$z_{1,2} = \frac32 \pm\sqrt{(\frac32)^2-2}$
$z_{1,2} = \frac32 \pm\sqrt{\frac14}$
$z_{1,2} = \frac32 \pm\frac12$

$z_1=2$ and $z_2=1$
3. #### Backward substitution

Now you can calculate $x$ from the solutions for $z$.
To do this, we take the original equation and transform it:

$x^2=z\quad|\pm\sqrt{}$
$x=\pm\sqrt{z}$

Use both z-values.
$x_{1,2}=\pm\sqrt{z_1}$
$x_1=\sqrt{2}\approx1.41$
$x_2=-\sqrt{2}\approx-1.41$

$x_{3,4}=\pm\sqrt{z_2}$
$x_3=\sqrt{1}=1$
$x_4=-\sqrt{1}=-1$