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Completing the square...
- Thread starter klooless
- Start date
Students often run into trouble when learning completing the square when the coefficient of the x^2 term is something other than 1.
No troubles. We have to take that into account.
\(\displaystyle 2x^{2}+2x=-5\)
\(\displaystyle 2(x^{2}+x)=-5\)
The coefficient of the x term is 1. Half of that is 1/2. Square that and we get 1/4.
2 times 1/4 is 1/2.
\(\displaystyle 2(x^{2}+x+\frac{1}{4})=-5+\frac{1}{2}\)
See there?. We had to multiply the term we added by 2.
\(\displaystyle 2(x+\frac{1}{2})^{2}+\frac{9}{2}\)
If you ever have trouble completing the square, just use the general form and plug in your a,b,c values.
If we have \(\displaystyle ax^{2}+bx+c\) and are asked to complete the square, use:
\(\displaystyle a(x+\frac{b}{2a})^{2}+\underbrace{c-\frac{b^{2}}{4a}}_{\text{constant term}}\)
No troubles. We have to take that into account.
\(\displaystyle 2x^{2}+2x=-5\)
\(\displaystyle 2(x^{2}+x)=-5\)
The coefficient of the x term is 1. Half of that is 1/2. Square that and we get 1/4.
2 times 1/4 is 1/2.
\(\displaystyle 2(x^{2}+x+\frac{1}{4})=-5+\frac{1}{2}\)
See there?. We had to multiply the term we added by 2.
\(\displaystyle 2(x+\frac{1}{2})^{2}+\frac{9}{2}\)
If you ever have trouble completing the square, just use the general form and plug in your a,b,c values.
If we have \(\displaystyle ax^{2}+bx+c\) and are asked to complete the square, use:
\(\displaystyle a(x+\frac{b}{2a})^{2}+\underbrace{c-\frac{b^{2}}{4a}}_{\text{constant term}}\)
Hello, klooless!
\(\displaystyle \text{Subtract 5 from both sides: }\;2x^2 + x \;=\;\text{-}5\)
\(\displaystyle \text{Divide by 2: }\;x^2 + x \;=\;\text{-}\tfrac{5}{2}\)
\(\displaystyle \text{Complete the square: }\;x^2 + x + {\bf\tfrac{1}{4}} \;=\;\text{-}\tfrac{5}{2} + {\bf\tfrac{1}{4}}\)
\(\displaystyle \text{Simplify: }\;\left(x + \tfrac{1}{2}\right)^2 \;=\;\text{-}\tfrac{9}{4}\)
\(\displaystyle \text{Take square roots: }\;x + \tfrac{1}{2} \;=\;\pm\sqrt{\text{-}\tfrac{9}{4}} \;=\;\pm\tfrac{3}{2}i\)
\(\displaystyle \text{Therefore: }\;\boxed{x \;=\;-\tfrac{1}{2} \pm\tfrac{3}{2}i}\)
I'm having trouble completing the square with this equation: .\(\displaystyle 2x^2 + 2x + 5 \:=\:0\)
\(\displaystyle \text{Subtract 5 from both sides: }\;2x^2 + x \;=\;\text{-}5\)
\(\displaystyle \text{Divide by 2: }\;x^2 + x \;=\;\text{-}\tfrac{5}{2}\)
\(\displaystyle \text{Complete the square: }\;x^2 + x + {\bf\tfrac{1}{4}} \;=\;\text{-}\tfrac{5}{2} + {\bf\tfrac{1}{4}}\)
\(\displaystyle \text{Simplify: }\;\left(x + \tfrac{1}{2}\right)^2 \;=\;\text{-}\tfrac{9}{4}\)
\(\displaystyle \text{Take square roots: }\;x + \tfrac{1}{2} \;=\;\pm\sqrt{\text{-}\tfrac{9}{4}} \;=\;\pm\tfrac{3}{2}i\)
\(\displaystyle \text{Therefore: }\;\boxed{x \;=\;-\tfrac{1}{2} \pm\tfrac{3}{2}i}\)
D