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Commentary for "Two methods for deriving the quadratic formula that I was not taught in school"

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Commentary for "Two methods for deriving the quadratic formula that I was not taught in school"

Moderator edit: This topic is for commentary pertaining to the tutorial:

Two methods for deriving the quadratic formula that I was not taught in school
Hi MarkFL, :)

To complete the square, I was taught to move the constant term to the other side and divide through by a:

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

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

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

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

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

Apply the square root property:

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

Solve for x:

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

And we have the famous quadratic formula.
We were taught(or maybe something that I picked up somewhere) to think about the completion of the square in the following manner.

\[ax^2+bx+c=0\Rightarrow x^2+\frac{b}{a}x+\frac{c}{a}=0\]

Now the first two terms could be made up by squaring \(x+\frac{b}{2a}\). But when we square it we get an additional term, \(\frac{b^2}{4a^2}\). Therefore,

\[x^2+ \frac{b}{a}x+ \frac{c}{a}=\left(x+\frac{b}{2a} \right)^2- \frac{b^2}{4a^2}+ \frac{c}{a}=0\]

Simplifying this we get the quadratic formula.

Method 2:

Arrange (1) as:

$\displaystyle ax^2+bx=-c$

Multiply by $\displaystyle 4a$:

$\displaystyle 4a^2x^2+4abx=-4ac$

Add $\displaystyle b^2$ to both sides:

$\displaystyle 4a^2x^2+4abx+b^2=b^2-4ac$

Write the left side as a square:

$\displaystyle (2ax+b)^2=b^2-4ac$

Apply the square root property:

$\displaystyle 2ax+b=\pm\sqrt{b^2-4ac}$

Solve for x:

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

I encountered this method when I was trying to solve quadratic congruences of prime moduli in number theory (see >>this<<).

Kind Regards,
Sudharaka.
 
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MarkFL

Administrator
Staff member
Feb 24, 2012
13,735
Re: Two methods for solving quadratic equations that I was not taught in school

That makes sense, as the second method allows you to work only with integers (until the last step, given that the coefficients of the quadratic are all integers.
 

earboth

Active member
Jan 30, 2012
74
Re: Two methods for solving quadratic equations that I was not taught in school

As a student, I was taught 3 ways to solve quadratic equations:

i) Factoring

ii) Completing the square

iii) Applying the quadratic formula, derived by completing the square on the general quadratic in standard form:

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

To complete the square, I was taught to move the constant term to the other side and divide through by a:

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

...
An additional remark only:

In Germany mostly the so-called pq-formula is used:

$\displaystyle ax^2+bx+c=0$

Norm this equation:

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

Rename the coefficients:

$\displaystyle x^2+px+q=0$

The solution is then:

$x = -\frac p2 \pm \sqrt{\frac{p^2}4-q}$

where $\frac{p^2}4 = \left(-\frac p2 \right)^2$
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,735
Re: Two methods for solving quadratic equations that I was not taught in school

I correspond with a student in Sweden who is used to the pq-formula.:)
 

mathmaniac

Active member
Mar 4, 2013
188
Re: Two methods for deriving the quadratic formula that I was not taught in school

I have a different method for solving the quadratic and deriving the formula....

Let the roots of the quadratic be \(\displaystyle ax^2+bx+c=0\) be $p$ and $q$.

Then \(\displaystyle (x-p)(x-q)=x^2-(p+q)x+pq=0\)

which gives \(\displaystyle p+q=-\frac{b}{a}\) and \(\displaystyle pq=\frac{c}{a}\)

Now we are left with solving these two above.....
Lets do it...

\(\displaystyle (p+q)^2- (p-q)^2=4pq\)

\(\displaystyle p-q=\sqrt{(p+q)^2​-4pq}\)

Equating \(\displaystyle p+q\) and \(\displaystyle p-q\) we get $p$ and $q$.

Now deriving the quadratic formula....

\(\displaystyle p-q=\left(-\frac{b}{a} \right)^2-\frac{4c}{a}=\frac{b^2-4ac}{a^2}\)

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

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

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

And Viola!!We got the quadratic formula

Somebody edit my latex...

Moderator edit: I fixed your $\LaTeX$ this time, but in the future please preview your posts and make sure they look the way you want before submitting the posts. You cannot expect the staff here to fix your posts on demand (especially when words like "please" and "thank you" are nowhere to be found).
 
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mathmaniac

Active member
Mar 4, 2013
188
Re: Two methods for deriving the quadratic formula that I was not taught in school

Sorry it was going difficult and time consuming for me,thats why I left it for you...
Thanks

Sorry for not including "please" and "thanks" but honestly,I always respect all who help me,especially you,Mark.....
 
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eddybob123

Active member
Aug 18, 2013
76
Re: Commentary for "Two methods for deriving the quadratic formula that I was not taught in school"

I've always wanted to figure out a geometric derivation of the quadratic formula. What I have tried probably isn't relevant. (Sun)
 

agentmulder

Active member
Feb 9, 2012
33
Re: Commentary for "Two methods for deriving the quadratic formula that I was not taught in school"

I've always wanted to figure out a geometric derivation of the quadratic formula. What I have tried probably isn't relevant. (Sun)
I'm not exactly sure what you are looking for but i can get an easy geometric derivation using MarkFL method 2 of the parent thread to this one.

http://mathhelpboards.com/pre-algeb...adratic-formula-i-not-taught-school-2629.html

Using

$ ax^2 + bx + c = 0 $

We consider 3 rectangles.

Rectangle 1 has side ax and x

Rectangle 2 has side b and x

Rectangle 3 has side 1 and c

Multiplying through by 4a can have the following interpretation (among other interpretations)

In Rectangle 1 , multiply side ax by 2 , multiply side x by 2a , this is the same as multiplying ax^2 by 4a and has turned Rectangle 1 into a geometric square with side 2ax , representing the (square) number $4a^2x^2 = (2ax)^2$

In Rectangle 2 , draw another copy , you now have 2 rectangles each with sides b and x , multiply each side x by 2a , so now you have two identical rectangles each with sides b and 2ax , the total is 2b*2ax = 4abx as it should be.

In Rectangle 3 , multiply side c by 4a , so now you have a rectangle with sides 1 and 4ac

The trick now is to take the two identical copies of Rectangle 2 and put them on the sides of square 2ax , the 'new square' will have sides 2ax + b and will be missing b^2 , so we give it a b^2 to complete it , then we subtract b^2 on the OUTSIDE of the completed square (2ax + b)^2 as a correction term.

So far it looks like this

$(2ax + b)^2 - b^2 + 4ac = 0$

The effect is you've enclosed the variable 'x' within a geometric square (or arithmetic square) and everything outside the square is a constant so you can go ahead and isolate the square , then extract the square root without fear and derive the Quadratic Formula using this 'geometric interpretation'.

I wish i had capability of showing diagrams as this is really easy to see by drawing the rectangles i've mentioned. I apologize if this is a bit confusing without diagrams but am supremely confident you can discover this yourself using one clean sheet of paper and one sharpened pencil by drawing the rectangles i've mentioned and following the procedure of method 2.

:D

[EDIT] Fixed a couple of typos and added a little more commentary [ENDEDIT]
 
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Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
Re: Commentary for "Two methods for deriving the quadratic formula that I was not taught in school"

Here is another derivation based on the notion that the graph of a quadratic function is a parabola:

We will suppose that our parabola is:

\(\displaystyle y = ax^2 + bx + c\).

For convenience, we will assume that \(\displaystyle a > 0\) (the more enterprising among you might want to investigate what changes when \(\displaystyle a < 0\), and why we do not consider \(\displaystyle a= 0\) at all).

We would like to put this parabola in "standard form":

\(\displaystyle y - k = A(x - h)^2\)

so that we can directly read off the location of its vertex at \(\displaystyle (h,k)\).

Rewriting:

\(\displaystyle y - k = A(x - h)^2\) as:

\(\displaystyle y = Ax^2 - 2Ahx + Ah^2 + k\), and comparing our two expressions, we see that:

\(\displaystyle A = a\)
\(\displaystyle -2Ah = b\)
\(\displaystyle Ah^2 + k = c\).

Solving for h and k in terms of a,b and c, we obtain:

\(\displaystyle h = \frac{b}{-2A} = -\frac{b}{2a}\)
\(\displaystyle k = c - Ah^2 = c - a\frac{b^2}{4a^2} = \frac{4ac - b^2}{4a}\).

Note this gives us an unexpected bonus: we know the parabola intersects the x-axis if and only if \(\displaystyle k \leq 0\), that is, if:

\(\displaystyle \frac{4ac - b^2}{4a} \leq 0\)

Which only happens when:

\(\displaystyle b^2 - 4ac \geq 0\) (since \(\displaystyle 4a > 0\)).

So now we can put our parabola in standard form as:

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

Since the roots of our quadratic equation are precisely the points on the graph of the parabola where y = 0, we set y equal to 0, and solve for x:


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

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

\(\displaystyle \frac{\pm\sqrt{b^2 - 4ac}}{2a} = x + \frac{b}{2a}\) (taking square roots of both sides)

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

Note that this involves no "guesswork" of deciding what we have to do to "complete the square", all we do is compare two polynomials term-by-term which we know are equal, and equate terms, and the completion is done for us (this is why we solve for h and k)!