Solve the equation x - 1/x = 3.

Step-by-step solution

1. 1

   Note the domain restriction before multiplying. The term $\tfrac1x$ requires $x\neq0$. Recording this now matters: multiplying through by $x$ is only reversible when $x\neq0$, and any root that turned out to be $0$ would have to be discarded as extraneous.

2. 2

   Clear the denominator. Multiply every term by $x$:

   $x\cdot x-x\cdot\frac{1}{x}=3x\qquad\Longrightarrow\qquad x^{2}-1=3x$

   The middle term becomes exactly $-1$, since $x\cdot\tfrac1x=1$. Every term must be multiplied, including the $3$ on the right.

3. 3

   Rearrange into standard form. Subtract $3x$ from both sides:

   $x^{2}-3x-1=0$

   so $a=1$, $b=-3$, $c=-1$.

4. 4

   Apply the quadratic formula.

   $x=\frac{3\pm\sqrt{(-3)^{2}-4(1)(-1)}}{2}=\frac{3\pm\sqrt{9+4}}{2}=\frac{3\pm\sqrt{13}}{2}$

   The discriminant $13$ is positive but not a perfect square, so there are two distinct irrational roots and no integer factorisation exists.

5. 5

   Check neither root is excluded. Numerically $x\approx3.3028$ and $x\approx-0.3028$; neither is $0$, so both survive the domain restriction. Substituting each into the original equation $x-\tfrac1x$ gives $3$ to within $10^{-12}$ ✓ — checking against the original, not the cleared version, is what catches extraneous roots.

6. 6

   Confirm with Vieta. The roots should sum to $-b/a=3$: $\dfrac{3+\sqrt{13}}{2}+\dfrac{3-\sqrt{13}}{2}=3$ ✓. They should multiply to $c/a=-1$: $\dfrac{9-13}{4}=-1$ ✓. The product being $-1$ also says the two roots are negative reciprocals of each other — a neat consequence of the equation's $x-\tfrac1x$ shape.