Solve the equation 2x^2 = 2x + 1.

Step-by-step solution

1. 1

   Put the equation in standard form. The quadratic formula needs $ax^{2}+bx+c=0$, so subtract $2x+1$ from both sides:

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

   giving $a=2$, $b=-2$, $c=-1$. Note $c$ is negative: moving $+1$ across changes its sign.

2. 2

   Compute the discriminant to choose a method.

   $\Delta=b^{2}-4ac=(-2)^{2}-4(2)(-1)=4+8=12$

   The double negative in $-4(2)(-1)=+8$ is the step to watch. Since $\Delta=12>0$ there are two distinct real roots, but $12$ is not a perfect square, so the roots are irrational and no integer factorisation exists — the formula is the right tool.

3. 3

   Substitute into the quadratic formula.

   $x=\frac{-b\pm\sqrt{\Delta}}{2a}=\frac{2\pm\sqrt{12}}{4}$

4. 4

   Simplify the surd and reduce the fraction. Since $12=4\cdot3$, $\sqrt{12}=2\sqrt{3}$:

   $x=\frac{2\pm2\sqrt{3}}{4}=\frac{2\left(1\pm\sqrt{3}\right)}{4}=\frac{1\pm\sqrt{3}}{2}$

   The common factor $2$ must be cancelled from both terms of the numerator — cancelling only the leading $2$ is a classic error.

5. 5

   Verify with Vieta and by substitution. The roots should sum to $-b/a=1$ and multiply to $c/a=-\tfrac12$:

   $\frac{1+\sqrt3}{2}+\frac{1-\sqrt3}{2}=1\ \checkmark,\qquad\frac{(1)^{2}-(\sqrt3)^{2}}{4}=\frac{1-3}{4}=-\frac12\ \checkmark$

   Numerically the roots are $\approx1.3660$ and $\approx-0.3660$, and substituting either into $2x^{2}-2x-1$ gives a residual below $10^{-12}$ ✓.