Fourth (?) Method to Solve a Quadratic

[JeffM]

In case any of you missed it, lookagain pointed out a new way to solve quadratics. It is really a variant of completing the square, but so is the quadratic formula.

\(\displaystyle ax^2 + bx + c = 0 \implies ax^2 + bx = - c.\) Looks familiar, but now multiply both sides of the equation by 4a.

\(\displaystyle 4a(ax^2 + bx) = 4a(-4c) \implies\)

\(\displaystyle 4a^2x^2 + 4abx = - 4ac.\) Now add b^2 to both sides.

\(\displaystyle 4a^2 + 4abx + b^2 = b^2 - 4ac \implies\)

\(\displaystyle (2ax + b)^2 = b^2 - 4ac \implies 2ax + b = \pm \sqrt{b^2 - 4ac} \implies x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\)

What is neat about this method is that, if the coefficients in the original quadratic are integers, the method involves no division or fractions until the last step. As a derivation of the quadratic formula, I prefer it because of its simplicity. Kids hate algebraic fractions.

 

[MarkFL]

Here is another method:

\(\displaystyle ax^2+bx+c=0\)

Divide by \(\displaystyle a\):

\(\displaystyle \displaystyle x^2+\frac{b}{a}x+\frac{c}{a}=0\)

Now, we next want to shift the roots to the right by 1/2 the value of the coefficient of the linear term, so our new equation is:

\(\displaystyle \displaystyle \left(x-\frac{b}{2a} \right)^2+\frac{b}{a}\left(x-\frac{b}{2a} \right)+\frac{c}{a}=0\)

\(\displaystyle \displaystyle x^2-\frac{b}{a}x+\frac{b^2}{4a^2}+\frac{b}{a}x-\frac{b^2}{2a^2}+\frac{c}{a}=0\)

\(\displaystyle \displaystyle x^2=\frac{b^2-4ac}{4a^2}\)

\(\displaystyle \displaystyle x=\pm\frac{\sqrt{b^2-4ac}}{2a}\)

Now, we subtract \(\displaystyle \displaystyle \frac{b}{2a}\) from these roots, to get the roots of the original equation:

\(\displaystyle \displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)

 

[JeffM]

Here is another method:

\(\displaystyle ax^2+bx+c=0\)

Divide by \(\displaystyle a\):

\(\displaystyle \displaystyle x^2+\frac{b}{a}x+\frac{c}{a}=0\)

Now, we next want to shift the roots to the right by 1/2 the value of the coefficient of the linear term, so our new equation is:

\(\displaystyle \displaystyle \left(x-\frac{b}{2a} \right)^2+\frac{b}{a}\left(x-\frac{b}{2a} \right)+\frac{c}{a}=0\)

\(\displaystyle \displaystyle x^2-\frac{b}{a}x+\frac{b^2}{4a^2}+\frac{b}{a}x-\frac{b^2}{2a^2}+\frac{c}{a}=0\)

\(\displaystyle \displaystyle x^2=\frac{b^2-4ac}{4a^2}\)

\(\displaystyle \displaystyle x=\pm\frac{\sqrt{b^2-4ac}}{2a}\)

Now, we subtract \(\displaystyle \displaystyle \frac{b}{2a}\) from these roots, to get the roots of the original equation:

\(\displaystyle \displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)

Yes that is the method of completing the square that I was taught some 55 years ago.

 

[MarkFL]

The method for completing the square I was taught in school was:

\(\displaystyle \displaystyle ax^2+bx+c=0\)

Move the constant term to the other side and divide through by a:

\(\displaystyle \displaystyle x^2+\frac{b}{a}x=-\frac{c}{a}\)

Then, add the square of one-half the coefficient of the linear term to both sides:

\(\displaystyle \displaystyle x^2+\frac{b}{a}x+\left(\frac{b}{2a} \right)^2=-\frac{c}{a}+\left(\frac{b}{2a} \right)^2\)

Write the left side as a square, and combine terms on the right:

\(\displaystyle \displaystyle \left(x+\frac{b}{2a} \right)^2=\frac{b^2-4ac}{(2a)^2}\)

Apply the square root property:

\(\displaystyle \displaystyle x+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a}\)

Solve for \(\displaystyle x\):

\(\displaystyle \displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)

 

[lookagain]

here is another method:

\(\displaystyle ax^2+bx+c=0\)

divide by \(\displaystyle a\):

\(\displaystyle \displaystyle x^2+\frac{b}{a}x+\frac{c}{a}=0 \ \ \) (1)

now, we next want to shift the roots to the right by 1/2 the value of the coefficient of the linear term, so our new equation is:

\(\displaystyle \displaystyle \left(x-\frac{b}{2a} \right)^2+\frac{b}{a}\left(x-\frac{b}{2a} \right)+\frac{c}{a}=0 \ \ \ \ \) <---- This isn't an equivalent equation to (1). If you want to make some shift, use some other variable, as in \(\displaystyle \ x \ = \ t \ - \dfrac{b}{2a}.\)

\(\displaystyle \displaystyle \left(t - \frac{b}{2a} \right)^2+\frac{b}{a}\left(t - \frac{b}{2a} \right)+\frac{c}{a}=0 \ \ \ \ \) <----- This would be the start instead.

\(\displaystyle \displaystyle t^2 -\frac{b}{a}t+\frac{b^2}{4a^2}+\frac{b}{a}t - \frac{b^2}{2a^2}+\frac{c}{a}=0\)

\(\displaystyle \displaystyle t^2=\frac{b^2-4ac}{4a^2}\)

\(\displaystyle \displaystyle t = \pm\frac{\sqrt{b^2-4ac}}{2a}\)

and because \(\displaystyle \ x \ = \ t - \dfrac{b}{2a}, \ \) then now we subtract \(\displaystyle \ \displaystyle \frac{b}{2a} \ \) from these roots, to get the roots of the original equation:

\(\displaystyle \displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)

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