Question

X^2-x-2=0 as a complete square

Answer 1

Answer:

$x_1=2; x_2=-1$

Step-by-step explanation:

The square term is $x^2$ so we expect the complete square to be of the form $(x-a) ^2=x^2+2ax+a^2$.

Let's compare the first degree terms now: we have $2ax$ on one side and $-1x$ on the other. That would make $a=- \frac12$. At this point we have still the second squared term, so we add and subtract $\frac14 = (\frac12)^2$

Our equation becomes:

$x^2-x+\frac14 -\frac14-2 =0\\(x-\frac12)^2-\frac94=0$

(the $\frac94$ term comes from rewriting $2= \frac84$ and adding the terms out of the bracket).

At this point it's just solving the equation:

$(x-\frac12)^2 = \frac94 \rightarrow x-\frac12=\pm\frac32\\x=\frac12\pm\frac32\\x_1=\frac42 = 2; x_2=\fra-\frac22=-1$