Answer:
[tex](x+\frac{5}{2})^2-\frac{1}{4}[/tex]
Explanation:
Given a quadratic equation of the form:
[tex]y=ax^2+bx+c[/tex]
To "complete the squares" is to rewrite it as:
[tex]y=a(x+d)^2+e[/tex]
Where:
[tex]d=\frac{b}{2a}[/tex][tex]e=c-\frac{b^2}{4a}[/tex]
In this case, we are given:
[tex]y=x^2+5x+6[/tex]
Then:
a = 1
b = 5
c = 6
Using the formulas above:
[tex]d=\frac{5}{2\cdot1}=\frac{5}{2}[/tex][tex]e=6-\frac{5^2}{4\cdot1}=6-\frac{25}{4}=-\frac{1}{4}[/tex]
Now, we can rewrite:
[tex]y=(x+\frac{5}{2})^2-\frac{1}{4}[/tex]
