need help completing the square
- Thread starter lil_Dk
- Start date
You can use different techniques. That 4 as the leading coefficient is tripping you up, ain't it?. They'll do that.
\(\displaystyle \L\\4x^{2}+3x+1=0\)
\(\displaystyle \L\\4x^{2}+3x=-1\)
\(\displaystyle \L\\4(x^{2}+\frac{3}{4}x)=-1\)
Now, take one-half the coefficient of x, square it, and add it inside the
parentheses;
take 4 times the number you just found and add it to the right side(-1).
one-half the coefficient of x is 3/8. The square of 3/8 is 9/64.
4 times 9/64 is 9/16.
\(\displaystyle \L\\4(x^{2}+\frac{3}{4}x+\frac{9}{64})=-1+\frac{9}{16}\)
Can you finish up?.
\(\displaystyle \L\\4x^{2}+3x+1=0\)
\(\displaystyle \L\\4x^{2}+3x=-1\)
\(\displaystyle \L\\4(x^{2}+\frac{3}{4}x)=-1\)
Now, take one-half the coefficient of x, square it, and add it inside the
parentheses;
take 4 times the number you just found and add it to the right side(-1).
one-half the coefficient of x is 3/8. The square of 3/8 is 9/64.
4 times 9/64 is 9/16.
\(\displaystyle \L\\4(x^{2}+\frac{3}{4}x+\frac{9}{64})=-1+\frac{9}{16}\)
Can you finish up?.
- Joined
- Feb 4, 2004
- Messages
- 15,943
Given ax<sup>2</sup> + bx + c = 0, the completing-the-square process is, step-by-step, as follows:
. . . . .Take the original equation:
. . . . .\(\displaystyle \large{ax^2\,+\,bx\,+\,c\,=\,0}\)
. . . . .Move the constant term to the other side:
. . . . .\(\displaystyle \large{ax^2\,+\,bx\,=\,-c}\)
. . . . .Factor out whatever is multiplied on the squared term:
. . . . .\(\displaystyle \large{a\left(x^2\,+\,\frac{b}{a}x\right)\,=\,-c}\)
. . . . .Take one-half of the coefficient of the linear term,
. . . . .square it, and add on both sides, taking account of
. . . . .anything you may have factored out:
. . . . .\(\displaystyle \large{\left(\frac{1}{2}\right)\,\left(\frac{b}{a}\right)\. =\, \frac{b}{2a}}\)
. . . . .\(\displaystyle \large{\left(\frac{b}{2a}\right)^2\, =\, \frac{b^2}{4a^2}}\)
. . . . .\(\displaystyle \large{a\left(x^2\,+\,\frac{b}{a}x\,+\,\frac{b}{4a^2}\right)\,=\,a\left(\frac{b}{4a^2}\right)\,-\,c \,=\,\frac{b}{4a}\,-\,c}\)
. . . . .Complete the square on the left-hand side:
. . . . .\(\displaystyle \large{a\left(x\,+\,\frac{b}{2a}\right)^2\,=\,\frac{b}{4a}\,-\,c}\)
. . . . .Divide through by whatever was factored out:
. . . . .\(\displaystyle \large{(x\,+\,\frac{b}{2a})^2\,=\,\frac{b}{4a^2}\,-\,\frac{c}{a}}\)
. . . . .Take the square root:
. . . . .\(\displaystyle \large{x\,+\,\frac{b}{2a}\,=\,\pm\,\sqrt{\frac{b}{4a^2}\,-\,\frac{c}{a}}}\)
. . . . .Move whatever is with the variable over onto the other side:
. . . . .\(\displaystyle \large{x\,= \,-\frac{b}{2a}\,\pm\,\sqrt{\frac{b}{4a^2}\,-\,\frac{c}{a}}}\)
So, for instance:
. . . . .\(\displaystyle \large{3x^2\, +\, 5x\, -\, 4\, =\, 0}\)
. . . . .\(\displaystyle \large{3x^2\,+\,5x\,=\,4}\)
. . . . .\(\displaystyle \large{3(x^2\,+\,\frac{5}{3}x)\,= \,4}\)
. . . . .\(\displaystyle \large{3(x^2\,+\,\frac{5}{3}x\,+\,\frac{25}{36})\, =\,4\,+\,3\left(\frac{25}{36}\right)\, =\,4\,+\,\frac{25}{12}\,=\,\frac{73}{12}}\)
. . . . .\(\displaystyle \large{3(x\,+\,\frac{5}{6})^2\,=\,\frac{73}{12}}\)
. . . . .\(\displaystyle \large{(x\,+\,\frac{5}{6})^2\,=\,\frac{73}{36}}\)
. . . . .\(\displaystyle \large{x\,+\,\frac{5}{6}\,= \,\pm\,\sqrt{\frac{73}{36}}}\)
. . . . .\(\displaystyle \large{x\,= \,-\frac{5}{6}\,\pm\,\sqrt{\frac{73}{36}}}\)
Simplify a bit, and you're done.
Your example works the same way.
Eliz.
Edit: Corrected typoes graciously pointed out by galactus. Thank you!
. . . . .Take the original equation:
. . . . .\(\displaystyle \large{ax^2\,+\,bx\,+\,c\,=\,0}\)
. . . . .Move the constant term to the other side:
. . . . .\(\displaystyle \large{ax^2\,+\,bx\,=\,-c}\)
. . . . .Factor out whatever is multiplied on the squared term:
. . . . .\(\displaystyle \large{a\left(x^2\,+\,\frac{b}{a}x\right)\,=\,-c}\)
. . . . .Take one-half of the coefficient of the linear term,
. . . . .square it, and add on both sides, taking account of
. . . . .anything you may have factored out:
. . . . .\(\displaystyle \large{\left(\frac{1}{2}\right)\,\left(\frac{b}{a}\right)\. =\, \frac{b}{2a}}\)
. . . . .\(\displaystyle \large{\left(\frac{b}{2a}\right)^2\, =\, \frac{b^2}{4a^2}}\)
. . . . .\(\displaystyle \large{a\left(x^2\,+\,\frac{b}{a}x\,+\,\frac{b}{4a^2}\right)\,=\,a\left(\frac{b}{4a^2}\right)\,-\,c \,=\,\frac{b}{4a}\,-\,c}\)
. . . . .Complete the square on the left-hand side:
. . . . .\(\displaystyle \large{a\left(x\,+\,\frac{b}{2a}\right)^2\,=\,\frac{b}{4a}\,-\,c}\)
. . . . .Divide through by whatever was factored out:
. . . . .\(\displaystyle \large{(x\,+\,\frac{b}{2a})^2\,=\,\frac{b}{4a^2}\,-\,\frac{c}{a}}\)
. . . . .Take the square root:
. . . . .\(\displaystyle \large{x\,+\,\frac{b}{2a}\,=\,\pm\,\sqrt{\frac{b}{4a^2}\,-\,\frac{c}{a}}}\)
. . . . .Move whatever is with the variable over onto the other side:
. . . . .\(\displaystyle \large{x\,= \,-\frac{b}{2a}\,\pm\,\sqrt{\frac{b}{4a^2}\,-\,\frac{c}{a}}}\)
So, for instance:
. . . . .\(\displaystyle \large{3x^2\, +\, 5x\, -\, 4\, =\, 0}\)
. . . . .\(\displaystyle \large{3x^2\,+\,5x\,=\,4}\)
. . . . .\(\displaystyle \large{3(x^2\,+\,\frac{5}{3}x)\,= \,4}\)
. . . . .\(\displaystyle \large{3(x^2\,+\,\frac{5}{3}x\,+\,\frac{25}{36})\, =\,4\,+\,3\left(\frac{25}{36}\right)\, =\,4\,+\,\frac{25}{12}\,=\,\frac{73}{12}}\)
. . . . .\(\displaystyle \large{3(x\,+\,\frac{5}{6})^2\,=\,\frac{73}{12}}\)
. . . . .\(\displaystyle \large{(x\,+\,\frac{5}{6})^2\,=\,\frac{73}{36}}\)
. . . . .\(\displaystyle \large{x\,+\,\frac{5}{6}\,= \,\pm\,\sqrt{\frac{73}{36}}}\)
. . . . .\(\displaystyle \large{x\,= \,-\frac{5}{6}\,\pm\,\sqrt{\frac{73}{36}}}\)
Simplify a bit, and you're done.
Your example works the same way.
Eliz.
Edit: Corrected typoes graciously pointed out by galactus. Thank you!
Hello, lil_Dk!
This quadratic has complex roots; I assume you're cool with that . . .
Divide by 4: \(\displaystyle \L\,g^2\,+\,\frac{3}{4}g\,+\,\frac{1}{4}\:=\:0\)
"Move" the constant term: \(\displaystyle \L\,g^2\,+\,\frac{3}{4}g\;=\;-\frac{1}{4}\)
Complete the square:
\(\displaystyle \;\;\)Take one-half of \(\displaystyle \frac{3}{4}\) and square: \(\displaystyle \,\left(\frac{1}{2}\cdot\frac{3}{4}\right)^2\,=\,\frac{9}{64}\)
Add to both sides: \(\displaystyle \L\,g^2\,+\,\frac{3}{4}g\,+\,\frac{9}{64}\;=\;-\frac{1}{4}\,+\,\frac{9}{64}\)
Simplify: \(\displaystyle \L\,\left(g\,+\,\frac{3}{8}\right)^2\:=\: -\frac{7}{64}\)
Take square roots: \(\displaystyle \L\,g\,+\,\frac{3}{8}\:=\:\pm\sqrt{-\frac{7}{64}}\:=\:\pm\frac{i\sqrt{7}}{8}\)
Therefore: \(\displaystyle \L\,g\:=\:-\frac{3}{8}\,\pm\,\frac{i\sqrt{7}}{8}\:=\:\frac{-3\,\pm\,i\sqrt{7}}{8}\)
This quadratic has complex roots; I assume you're cool with that . . .
I prefer to get rid of the leading coefficient first . . .
Divide by 4: \(\displaystyle \L\,g^2\,+\,\frac{3}{4}g\,+\,\frac{1}{4}\:=\:0\)
"Move" the constant term: \(\displaystyle \L\,g^2\,+\,\frac{3}{4}g\;=\;-\frac{1}{4}\)
Complete the square:
\(\displaystyle \;\;\)Take one-half of \(\displaystyle \frac{3}{4}\) and square: \(\displaystyle \,\left(\frac{1}{2}\cdot\frac{3}{4}\right)^2\,=\,\frac{9}{64}\)
Add to both sides: \(\displaystyle \L\,g^2\,+\,\frac{3}{4}g\,+\,\frac{9}{64}\;=\;-\frac{1}{4}\,+\,\frac{9}{64}\)
Simplify: \(\displaystyle \L\,\left(g\,+\,\frac{3}{8}\right)^2\:=\: -\frac{7}{64}\)
Take square roots: \(\displaystyle \L\,g\,+\,\frac{3}{8}\:=\:\pm\sqrt{-\frac{7}{64}}\:=\:\pm\frac{i\sqrt{7}}{8}\)
Therefore: \(\displaystyle \L\,g\:=\:-\frac{3}{8}\,\pm\,\frac{i\sqrt{7}}{8}\:=\:\frac{-3\,\pm\,i\sqrt{7}}{8}\)