Completing the square allows us to convert and expression like
[tex]x^2+bx[/tex]into a square of the form
[tex](x+c)^2+e.[/tex]To apply this method, we need to recall first the following identity
[tex](y+z)^2=y^2+2yz+z^2.[/tex]Our task is to take x^2+x and convert it into an expression that looks like the right-hand side of the identity above. On the left, we have x squared; we should then consider x as y in the identity above.
After this square, we have a single x; but we want something like "2yz" there. We only have "y" so far. Multiplying and dividing by 2, we get
[tex]x=2\cdot(\frac{1}{2})\cdot y=2y(\frac{1}{2})\text{.}[/tex]if we set z=1/2, we're done.
Finally, we need to add z^2, which in this case is
[tex](\frac{1}{2})^2=\frac{1}{4}\text{.}[/tex]To avoid affecting the expression, we must not only add z^2 but subtract it as well.
In summary, we get
[tex]x^2+x=x^2+2x(\frac{1}{2})+(\frac{1}{2})^2-(\frac{1}{2})^2\text{.}[/tex]And applying the identity, we obtain
[tex]x^2+x=(x+\frac{1}{2})^2-\frac{1}{4}.[/tex]Then, the equation of exercise turns out to be
[tex](x+\frac{1}{2})^2-\frac{1}{4}=\frac{19}{4}\text{.}[/tex]Let's solve it:
[tex]\begin{gathered} (x+\frac{1}{2})^2-\frac{1}{4}=\frac{19}{4}, \\ \\ (x+\frac{1}{2})^2=\frac{19}{4}+\frac{1}{4}, \\ \\ (x+\frac{1}{2})^2=\frac{20}{4}, \\ \\ (x+\frac{1}{2})^2=5, \\ \\ \sqrt[]{(x+\frac{1}{2})^2}=\sqrt[]{5},\leftarrow\text{ Taking square root on both sides} \\ \\ |x+\frac{1}{2}|=\sqrt[]{5},\leftarrow\text{ The square root of a square is the absolute value of what is within the square} \end{gathered}[/tex]Let's solve the absolute value equation:
AnswerThe equation of the exercise has two solutions:
[tex]x=\frac{-1+2\cdot\sqrt[]{5}}{2},x=\frac{-1-2\cdot\sqrt[]{5}}{2}\text{.}[/tex]
