# Evaluating a fraction divided by a fraction

#### fortheentries

##### New member
Hi everyone. I'm just revising some quadratics in preparation for tutoring high school maths and came across something about fractions that has made me question whether I've been learning maths wrong my entire life. I was solving a quadratic by completing the square for this question:
$x^2-(3/2)x=0$
Obviously to complete the square I would have to add:
$(\frac{b}{2})^2$
Subbing in b I got:
$x^2-(3/2)x+(\frac{\frac{3}{2}}{2})^2= (\frac{\frac{3}{2}}{2})^2$
I've always thought for this type of issue where you have a fraction div by a fraction you multiply by the reciprocal so to evaluate the b/2 section:

$(\frac{3}{2}*\frac{2}{1})^2 \\ (\frac{6}{2})^2\\ 3^2= 9$However, the worked answers display the b/2 section working as
$(\frac{3}{2*2})^2$
Which is correct?

Hi everyone. I'm just revising some quadratics in preparation for tutoring high school maths and came across something about fractions that has made me question whether I've been learning maths wrong my entire life. I was solving a quadratic by completing the square for this question:
$x^2-(3/2)x=0$
Obviously to complete the square I would have to add:
$(\frac{b}{2})^2$
Subbing in b I got:
$x^2-(3/2)x+(\frac{\frac{3}{2}}{2})^2= (\frac{\frac{3}{2}}{2})^2$
Yes, you divide by flipping and multiplying, but you didn't flip the 2 = 2/1 to be multiplying by 1/2.

Eliz.

Hi everyone. I'm just revising some quadratics in preparation for tutoring high school maths and came across something about fractions that has made me question whether I've been learning maths wrong my entire life. I was solving a quadratic by completing the square for this question:
$x^2-(3/2)x=0$
If I were giving advice on this I would rewrite it as: [imath]2x^2-3x=0[/imath].
Which in turn gives
[imath]x(2x-3)=0\\x=0~\vee~x=\dfrac{3}{2}[/imath].

I would rewrite it [to avoid working with fractions]
You mean that's how you would type it into wolframalpha.

I might, also, but the topic of this discussion is how to work with ratios, at the high-school level. Personally, I'm confident that more than 50% of high school students enrolling in college need to revisit fractions. High school students ought not be afforded a pass.

[imath]\;[/imath]

I might, also, but the topic of this discussion is how to work with ratios, at the high-school level. Personally, I'm confident that more than 50% of high school students enrolling in college need to revisit fractions. High school students ought not be afforded a pass.
Actually, since the problem includes a fraction, you can't avoid working with fractions in solving it; but you can decide how you want to do so.

Multiplying through by 2, so that you find [imath]2\cdot\frac{3}{2}=3[/imath], is a little easier than the division (though many students have trouble even with this). In fact, I often suggest avoiding division, if only because writing a fraction on top of a fraction is confusing, and risks exactly the error that was made here, even if in principle you know what to do. So, for example, I recommend clearing fractions before solving (which is what this is), which still gives practice working with fractions.

In addition, though, solving by completing the square when there are only two terms is not the best strategy, and it's good to be aware of that, even though practice with both methods is a good idea. I'll often commend a student for any correct work they did while taking the hard way, and then show how they could have avoided it.

But of course you are right that very many students (even in calculus!) struggle with fractions, and need to practice doing all the operations correctly. I've seen it constantly. Both fractions and completing the square are worth learning well; but avoiding them when there are easier and safer methods is also worth learning. (I'd be interested to know whether the instructions for the problem specified completing the square.)

Why would you complete the square for this problem??

Whenever you have x^2 + bx =0, you should factor out the x.

This give you x(x+b) = 0 or x=0 or x=-b.

If you have any number and divide it by 2 and divide that result by 2, you are dividing the original number by 4 (2*2).

In fact, if you take any number, divide it by a and divide that result by b, you are basically dividing the original number by ab.

For example, suppose you want to compute $$\displaystyle \dfrac{72}{6} = \dfrac{72}{2*3}$$.
One way of computing this is to divide 72 by 2, get 36, and divide that by 3 (this method is known as reducing). That is. you are computing $$\displaystyle \dfrac{\frac{72}{2}}{3}$$

One more thing. You should know that dividing a number by 2 or multiply that number by 2 almost always give different results. When will you get the same results?