Solve the equation x^2 - 2x - 4 = 0.

Step-by-step solution

1. 1

   Check for a factorisation first. We need integers with product $-4$ and sum $-2$: the pairs $(1,-4)$ and $(2,-2)$ and $(4,-1)$ give sums $-3$, $0$ and $3$. None works, so use the formula with $a=1$, $b=-2$, $c=-4$.

2. 2

   Compute the discriminant.

   $\Delta=(-2)^2-4(1)(-4)=4+16=20$

   Positive, so two distinct real roots.

3. 3

   Substitute into the quadratic formula.

   $x=\frac{2\pm\sqrt{20}}{2}$

4. 4

   Simplify the surd and cancel. $20=4\cdot 5$, so $\sqrt{20}=2\sqrt5$ and every term carries a factor $2$:

   $x=\frac{2\pm 2\sqrt5}{2}=1\pm\sqrt5$

   Cancelling before simplifying the radical would be wrong: $\tfrac{2\pm\sqrt{20}}{2}\neq 1\pm\sqrt{20}$, because the $2$ must divide both numerator terms.

5. 5

   Check by completing the square. $x^2-2x-4=(x-1)^2-5$, which vanishes exactly when $(x-1)^2=5$, i.e. $x=1\pm\sqrt5$ $\checkmark$. Numerically $x\approx 3.2361$ or $-1.2361$.