1
$\begingroup$

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.

Equation 2 is the product of Equation 1 times 7. Apparently 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?

$\endgroup$
  • $\begingroup$ In most cases the square root cannot be evaluated to rational values. $\endgroup$ – user202729 Dec 12 '17 at 13:07
  • $\begingroup$ The common factor in the first example is $2$, not $4$. $\endgroup$ – B. Goddard Dec 12 '17 at 13:10
  • $\begingroup$ Your question will be easier to read, particularly the exponents, if you use mathjax math.meta.stackexchange.com/questions/5020/… $\endgroup$ – Lee Mosher Dec 12 '17 at 13:14
0
$\begingroup$

Suppose that we have some quadratic $q(x) = Ax^2 + Bx + C$, where $A, B, C$ are integers with a common factor $k$. Since they have a common factor $k$, there exist integers $a, b, c$ such that $A = ak$, $B = bk$, $C = ck$. Applying the quadratic formula to $q(x)$ gives $$ \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} = \frac{-bk \pm \sqrt{k^2b^2 - 4k^2ac}}{2ka} = \frac{-bk \pm k\sqrt{b^2 - 4ac}}{2ka}$$ and so we can see that the common factor $k$ will definitely show up in both the numerator and denominator of the roots given by the quadratic formula.

However, you may get "false positives" this way. For example, consider $q(x) = x^2 + 2x + 1$. Then, the quadratic formula gives the roots as $$\frac{-2 \pm \sqrt{4 - 4}}{2} = -\frac{2}{2}$$ but $2$ is not a common divisor of $a, b, c$.

$\endgroup$
  • $\begingroup$ Let's consider Class 2, the class of expressions like yours, expressions for which 2 is the factor in reducing the fraction but not a common common divisor of a,b,c. Will Class 2 expressions have anything interesting in common? $\endgroup$ – Chaim Dec 12 '17 at 15:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.