Solve 2(x² + 1/x²) − 11(x − 1/x) + 8 = 0.

Step-by-step solution

1. 1

   Choose the substitution that matches the bracket present. The equation contains $x-\tfrac1x$, so set

   $t=x-\frac1x\qquad(x\neq 0)$

   Squaring gives $t^{2}=x^{2}-2+\tfrac{1}{x^{2}}$, hence

   $x^{2}+\frac{1}{x^{2}}=t^{2}+2$

   Note the plus $2$ here. With the other common substitution $t=x+\tfrac1x$ the constant would be $-2$; mixing the two up is the classic error.

2. 2

   Rewrite the equation in t.

   $2\left(t^{2}+2\right)-11t+8=0\quad\Longrightarrow\quad 2t^{2}-11t+12=0$

3. 3

   Solve the quadratic in t.

   $t=\frac{11\pm\sqrt{121-96}}{4}=\frac{11\pm 5}{4}\quad\Longrightarrow\quad t=4\ \text{or}\ t=\frac32$

4. 4

   Unwind t = 4.

   $x-\frac1x=4\ \Longrightarrow\ x^{2}-4x-1=0\ \Longrightarrow\ x=2\pm\sqrt5$

   Both are real, since the discriminant is $16+4=20>0$.

5. 5

   Unwind t = 3/2.

   $x-\frac1x=\frac32\ \Longrightarrow\ 2x^{2}-3x-2=0\ \Longrightarrow\ (2x+1)(x-2)=0\ \Longrightarrow\ x=2\ \text{or}\ x=-\frac12$

   $\boxed{x=2+\sqrt5,\ 2-\sqrt5,\ 2,\ -\tfrac12}$

6. 6

   Note why all four survive, and check one. Unlike the companion equation with $t=x+\tfrac1x$ (where $|t|\ge 2$ rejects some branches), $t=x-\tfrac1x$ takes every real value, so neither branch is excluded. Checking $x=2$: $x^{2}+\tfrac{1}{x^{2}}=4.25$ and $x-\tfrac1x=1.5$, giving $2(4.25)-11(1.5)+8=8.5-16.5+8=0$ ✓.