Solve the equation x^4 - 6x^2 + 8 = 0.

Step-by-step solution

1. 1

   Recognise the quadratic pattern in disguise. Only even powers of $x$ appear, and $x^4=(x^2)^2$. Setting

   $u=x^2$

   turns the quartic into an ordinary quadratic — this shape is called a biquadratic.

2. 2

   Solve the quadratic in $u$. We need two numbers with product $8$ and sum $-6$, namely $-2$ and $-4$:

   $u^2-6u+8=(u-2)(u-4)=0\quad\Longrightarrow\quad u=2\ \text{or}\ u=4$

3. 3

   Substitute back and take roots. Each positive value of $u$ contributes two values of $x$:

   $x^2=2\Rightarrow x=\pm\sqrt2,\qquad x^2=4\Rightarrow x=\pm 2$

   Both $u$ values are positive, which is why all four roots are real. A negative $u$ would have produced imaginary roots instead.

4. 4

   Check one irrational and one integer root. At $x=\sqrt2$: $x^4=4$, $6x^2=12$, so $4-12+8=0$ $\checkmark$. At $x=2$: $16-24+8=0$ $\checkmark$.

5. 5

   Confirm the count. A degree-4 polynomial has at most four roots, and the factorisation

   $x^4-6x^2+8=(x^2-2)(x^2-4)=(x-\sqrt2)(x+\sqrt2)(x-2)(x+2)$

   exhibits all four, so nothing is missing.