Solve 7(x + 1/x) − 2(x² + 1/x²) = 9.

Step-by-step solution

1. 1

   Recognise the reciprocal symmetry. The equation is unchanged if $x$ is replaced by $\tfrac1x$, which is the signal to substitute

   $t=x+\frac1x$

   Note $x=0$ is excluded from the domain from the outset.

2. 2

   Express the second bracket in terms of t. Squaring the substitution:

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

   The cross term contributing the $+2$ is the detail that is easiest to drop.

3. 3

   Rewrite and tidy the equation.

   $7t-2\left(t^{2}-2\right)=9\ \Longrightarrow\ 7t-2t^{2}+4=9\ \Longrightarrow\ 2t^{2}-7t+5=0$

4. 4

   Solve for t.

   $(2t-5)(t-1)=0\quad\Longrightarrow\quad t=\frac52\ \text{or}\ t=1$

5. 5

   Unwind each value of t. For $t=\tfrac52$:

   $x+\frac1x=\frac52\ \Longrightarrow\ 2x^{2}-5x+2=0\ \Longrightarrow\ (2x-1)(x-2)=0\ \Longrightarrow\ x=\tfrac12\ \text{or}\ x=2$

   For $t=1$:

   $x+\frac1x=1\ \Longrightarrow\ x^{2}-x+1=0$

   whose discriminant is $1-4=-3<0$: no real solutions. This rejection is forced, since for real $x\neq 0$ the quantity $x+\tfrac1x$ always satisfies $|t|\ge 2$, and $t=1$ violates that.

   $\boxed{x=2\ \text{or}\ x=\tfrac12}$

6. 6

   Check both roots. At $x=2$: $x+\tfrac1x=2.5$ and $x^{2}+\tfrac{1}{x^{2}}=4.25$, so $7(2.5)-2(4.25)=17.5-8.5=9$ ✓. At $x=\tfrac12$ the same two quantities take the same values (reciprocal symmetry), so it checks identically ✓.