Canceling in fractions sometimes gives a wrong result

Question

When the same factor appears in the numerator and denominator, it can be canceled out: $$\frac{4a}{4a} = 1$$ However, in this more complicated fraction this does not work: $$\frac{4ac-b^2}{4a} \neq c-b^2$$ Why is that?

Answer 1

$$\frac{4ac-b^2}{4a} = \frac{4ac}{4a}-\frac{b^2}{4a} = c-\frac{b^2}{4a} \neq c-b^2 \textrm{ if } a \neq \frac14$$

Answer 2

Assuming you want to know when

$$a(x-h)^2+k=ax^2+bx+c\iff ax^2-2ahx+ah^2+k=ax^2+bx+c\iff$$

$$\begin{cases}-2ah=b\iff a=-\frac b{2h}\\{}\\ah^2+k=c\iff a=\frac{c-k}{h^2}\end{cases}$$

Comparing the above we get

$$-\frac b{2h}=\frac{c-k}{h^2}\iff-bh=2c-2k$$

Without any more information I don't think one can say much more...

Answer 3

It seems that the core of your issue is "cancellation." You can only "cancel" common factors of the numerator and denominator. Otherwise, you could pull nonsense like $$\frac32=\frac{2+1}2=1$$ by exactly the same reasoning as in your post (after all, isn't $\frac22=1$?)

When we do "cancellation" with fractions, here's what's really going on behind the scenes. If $x,y,z$ are numbers with $y,z\ne0,$ then $$\frac{xz}{yz}=\frac{x\cdot z}{y\cdot z}=\frac xy\cdot\frac zz=\frac xy\cdot 1=\frac xy,$$ or for a direct numerical example, $$\frac{6}{15}=\frac{2\cdot 3}{5\cdot 3}=\frac25\cdot\frac33=\frac25\cdot 1=\frac25.$$ We don't usually show all of these steps, but it's good to keep in mind that we're doing this. Let's look at another example: $$\frac{xz+z}{z}=\frac{(x+1)\cdot z}{1\cdot z}=\frac{x+1}1\cdot\frac zz=(x+1)\cdot 1=x+1.$$ It's tempting to just try to "cancel" by crossing out the $z$ on top and bottom (to get $xz$), but you must keep in mind what we're actually doing, because with that kind of reasoning you get (for example) that $$\frac{6+2}{2}``="6,$$ when what we should be getting is $$\frac{6+2}{2}=\frac{(3+1)\cdot 2}{1\cdot 2}=\frac{3+1}1\cdot\frac22=(3+1)\cdot 1=3+1=4$$ (which could be done more simply, but I wanted to illustrate the process again).

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Regarding the application (to the quadratic equation) in question, we can still "cancel", but we have to be a bit careful, and introduce the factor $4a$ in a clever way, using the fact that $a\ne 0$: $$\frac{4ac-b^2}{4a}=\frac{4ac-1\cdot b^2}{4a}=\cfrac{4ac-\frac{4a}{4a}b^2}{4a}=\cfrac{\left(c-\frac1{4a}b^2\right)\cdot 4a}{4a}=\left(c-\frac1{4a}b^2\right)=c-\frac{b^2}{4a}.$$

Answer 4

Your question is "Why can't I cancel the $4a$ terms in the top and bottom of a fraction to get $$ \frac{\color{red}{4a}c-b^2}{\color{red}{4a}}=c-b^2 $$ The problem with this is that on the left side, the term $4ac-b^2$ must be divided entirely by $4a$. You've only divided part of the top by $4a$. In other words, to get each term on the top to be a multiple of $4a$ we could multiply the $b^2$ term by $4a/4a$ to get $$ \frac{4ac-b^2}{4a}=\frac{4ac-\frac{4a}{4a}\cdot b^2}{4a}=\frac{4a\cdot c-4a\cdot \frac{b^2}{4a}}{4a}=\frac{4a(c-\frac{b^2}{4a})}{4a} $$ and now you can indeed do the cancellation of the $4a$ terms to conclude $$ \frac{4ac-b^2}{4a}=c-\frac{b^2}{4a} $$

Hope this helps.