Solve x⁴ − 24x² − 25 = 0.

Step-by-step solution

1. 1

   Recognise the biquadratic pattern. Only even powers of $x$ appear, so the substitution

   $t=x^{2}\quad(\text{with }t\ge 0\text{ for real }x)$

   turns a quartic into a quadratic. Recording the constraint $t\ge 0$ now is what makes the later rejection automatic.

2. 2

   Rewrite and solve the quadratic in t.

   $t^{2}-24t-25=0\quad\Longrightarrow\quad (t-25)(t+1)=0$

   (The pair with product $-25$ and sum $-24$ is $-25$ and $1$.) So $t=25$ or $t=-1$.

3. 3

   Reject the inadmissible value. $t=x^{2}$ cannot be negative for real $x$, so $t=-1$ contributes no real solutions.

4. 4

   Unwind the admissible value.

   $x^{2}=25\quad\Longrightarrow\quad x=\pm 5$

   $\boxed{x=5\ \text{or}\ x=-5}$

5. 5

   Check both roots. At $x=5$: $625-24(25)-25=625-600-25=0$ ✓. At $x=-5$ the same, since only even powers occur ✓.

6. 6

   Note the complete factorisation. $x^{4}-24x^{2}-25=\left(x^{2}-25\right)\left(x^{2}+1\right)=(x-5)(x+5)\left(x^{2}+1\right)$. The quartic has four roots in total — $\pm 5$ real and $\pm i$ complex — which is worth stating if the question asks for all roots rather than the real ones.