Solve 25x^4 + 10x^2 - 4 = 0.

Step-by-step solution

1. 1

   Recognise the biquadratic form. Only even powers of $x$ appear, so the substitution $u=x^{2}$ turns a quartic into a quadratic: $25u^{2}+10u-4=0.$ This works for any $ax^{4}+bx^{2}+c$ and is far quicker than the general quartic formula.

2. 2

   Solve the quadratic in $u$. With $a=25$, $b=10$, $c=-4$, $u=\frac{-10\pm\sqrt{100-4(25)(-4)}}{50}=\frac{-10\pm\sqrt{500}}{50}=\frac{-10\pm 10\sqrt5}{50}=\frac{-1\pm\sqrt5}{5}.$ Note $\sqrt{500}=10\sqrt5$, and the common factor $10$ cancels against the $50$.

3. 3

   Undo the substitution, one case at a time. With $u=x^{2}$ each value of $u$ contributes two values of $x$ - but only if $u\ge 0$. Since $\sqrt5=2.2360680$, $u_{1}=\frac{\sqrt5-1}{5}=0.2472136>0,\qquad u_{2}=\frac{-1-\sqrt5}{5}=-0.6472136<0.$

4. 4

   Extract the real solutions from the positive root. $x^{2}=\frac{\sqrt5-1}{5}\quad\Longrightarrow\quad x=\pm\sqrt{\frac{\sqrt5-1}{5}}=\pm 0.4972055.$ These are the only real solutions, so the quartic's graph crosses the axis exactly twice.

5. 5

   Extract the imaginary solutions from the negative root. $x^{2}=-\frac{1+\sqrt5}{5}\quad\Longrightarrow\quad x=\pm i\sqrt{\frac{1+\sqrt5}{5}}=\pm 0.8044959\,i.$ Forgetting to test the sign of $u$ before taking a square root is the standard error here, and it either invents two false real roots or loses two complex ones.

6. 6

   Verify by substitution. At $x=0.4972055$: $x^{2}=0.2472136$ and $x^{4}=0.0611145$, so $25(0.0611145)+10(0.2472136)-4=1.5278640+2.4721360-4=0.0000000$. As a second check, the four roots' product should be $c/a=-4/25=-0.16$, and $\left(0.4972055\right)^{2}\times\left(-\left(0.8044959\right)^{2}\right)=0.2472136\times(-0.6472136)=-0.1600000$.