In the final throes of the quadratic formula, you reduce a fraction. Consider the following two examples.

  1. $y = 6x^2 + 11x + 3$; the quadratic formula reveals the roots $x = -4/12$ or $x = -18/12$. The parts of the first fraction (4 and 12) have a common factor of 4, so the fraction reduces to -1/3. The parts of the second fraction (18 and 12) have a common factor of 6, so the fraction reduces to -3/2. The common factor of all three parts (4, 12 and 18) is 2.

  2. $y = 42x^2 + 77x + 21$; the quadratic formula reveals the roots $x = -28/84$ or $x = -126/84$. The parts of the first fraction (28 and 84) have a common factor of 28, so the fraction reduces to -1/3. The parts of the second fraction (126 and 84) have a common factor of 42, so the fraction reduces to -3/2. The common factor of all three parts (28, 84 and 126) is 14.

Now. Compare the two solutions.

We multiplied by 7 the right side of Equation 1 to produce Equation 2. As a result, the common factors of the equations were all multiplied by 7, and after reducing the fractions we discover the same roots.

My question is whether that step, reducing the fraction, reveals anything about the numerical factors of the equations.

It seems to me that any common numerical factor of a, b and c (like 7 in Equation 2) will also be a common factor of the parts of the fractions, and cancel when we reduce the fraction. But the parts of the fraction can have other common factors (like 2 in both Equations).

So when we come to that step, of reducing the fraction, does the common factor reveal anything about the numerical common factors of a, b and c?

  • 4
    $\begingroup$ This question might be better suited to MathSE since it is about the mathematics of the quadratic formula and not really about teaching it, but MathSE might not like this question as much as I like it. $\endgroup$ Dec 12, 2017 at 16:00
  • $\begingroup$ This is a great question, because I think it can look at the nature of the solutions for when we deal with higher order polynomials and you have me rethinking a few lecture notes. $\endgroup$ Dec 12, 2017 at 17:06
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    $\begingroup$ This is a poorly stated and very elementary question about solving quadratic equations that belongs on MathSE. It does not address (explicitly at least) how to teach solving quadratic equations. Anyone teaching the solution of such equations should be able to answer the questions asked. It is poorly stated - a,b,c are nowhere defined, the author speaks of finding the roots of $y = f(x)$, etc. $\endgroup$
    – Dan Fox
    Dec 13, 2017 at 8:49
  • $\begingroup$ @DanFox I don't think the OP needs to define a, b, ,c since he talked about the quadratic formula. The variables a, b, and c are clearly referring to the quadratic formula. $\endgroup$
    – Amy B
    Dec 15, 2017 at 6:38

2 Answers 2


This would be a good process to introduce the rational root test in quadratic form. Seeing rational answers and comparing to the possible roots from the rational root test. Thus when you extend using the rational root test to higher order polynomials the seeds of the idea would already be there.

Using your example the rational root test says the possible solutions are:

Factors of $c = \pm 1, \pm 3$

Factor of $a = \pm 1, \pm 2, \pm 3, \pm 6$

Possible rational solutions are: $\pm \frac{1}{6}, \pm \frac{1}{3}, \pm \frac{1}{2}, \pm 1, \pm \frac{3}{2}, \pm 3$

The solutions are there and the you can see the source of the common factors.

Also you can illustrate the limits of the rational root test by introducing irrational answers that don't meet the test.

Example: $ x^2+8x-7=0$

The rational root test says this would have rational roots $\pm 1, \pm 7$

The quadratic formula gives $-4\pm\sqrt{23} \approx 0.796 \text{and} -8.796$. Looking at the rational possible roots of $1$ and $-7$ we see it is near, relatively. Which would be an easy entry to some of the approximation techniques on polynomials.

I like this as an approach to add earlier in my presentation of quadratics to prepare for higher order polynomials solutions. I am adding that into my notes for the quadratic formula for next year.


Let $y=a d x^2+ bd x + cd$ with $a,b,c,d \in \mathbb{Z}$ with $a,d \neq 0$, i.e., suppose the coefficients of the quadratic share a common factor of $d$. Then \begin{align*} x &= \frac{ - bd \pm \sqrt{(bd)^2 - 4 ad cd}}{2 ad} \\ &= \frac{-bd \pm \sqrt{d^2 (b^2 - 4ac)}}{d2a} \\ &= \frac{-bd \pm |d|\sqrt{b^2-4ac}}{d2a} \tag{1}. \end{align*}

There are two possibilities: either $d >0 $ or $d<0$. If $d>0$, equation (1) reduces to $$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$ and we have the exact same roots as $y=ax^2 + bx + c$.

If $d<0$, then $|d| = - d$. Using this in equation (1) gives \begin{align*} x&= \frac{-bd \pm -d\sqrt{b^2-4ac}}{d2a} \\ &= \frac{-b \mp \sqrt{b^2 - 4ac}}{2a} \\ &= \frac{-b \pm \sqrt{b^2-4ac}}{2a} \end{align*} and again we have the same roots as $y=ax^2 + bx + c$. Thus rescaling all the coefficients doesn't change the roots, although it should be noted that this isn't necessarily the only cancellation that can happen when simplifying the roots!

An easier way to see this result is to recall the fundamental theorem of algebra, wherein any quadratic can be uniquely factored (up to ordering) as $$y= a (x- r_1)(x - r_2)$$ where $r_1, r_2$ are the roots. Changing the factor of $a$ to $a'$ doesn't affect the roots.


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