# Solving this simple fraction

$$-\frac{a}{5}\left(2\times\frac{a^2}{5}\right)^2+5\left(-\frac{a}{5}\right)^3\left(2\times\frac{a^2}{5}\right) -a^2\left(-\frac{a}{5}\right)\left(2\times\frac{a^2}{5}\right)=0$$

I get

$$\frac{-4a^5}{25}-\frac{2a^5}{125}+\frac{2a^5}{25}$$

Where I multiply the first and third fraction by 5 in the numerator and denominator to get 125 in the denominator

For some reason, I get the wrong answer of

$$-\frac{12a^5}{125}$$

$$\frac{4a^5}{125}$$

What have I done wrong?

• Your first term should be $\frac{-4a^5}{125}$ not $\frac{-4a^5}{25}$. – Olivier Oloa Nov 27 '16 at 14:16
• Your first term is all squared, including the 5 in the denominator... so you should have 125 in the first terms denominator – Eleven-Eleven Nov 27 '16 at 14:16
• The first problem that I see in your question, is that you start off with an equation, and end up with an expression. A solution to your equation should be a constant value (or several constant values). It's a little hard to see how the solution to the given equation could be either one of the non-constant expressions that you have noted ($-\frac{12a^5}{125}$ or $\frac{4a^5}{125}$). – barak manos Nov 27 '16 at 14:29
• Thanks! I feel dumb now... – Mataunited17 Nov 27 '16 at 14:59

we have $$(2\frac{a^2}{5})\left(-\frac{2a^3}{25}-\frac{a^3}{25}+\frac{a^3}{5}\right)$$=$$\frac{2a^2}{5}\cdot \frac{2a^3}{25}=$$ $$\frac{4a^5}{125}=0$$ thus $$a=0$$

• Does this really need an explanation? – Rohan Nov 27 '16 at 14:25
• those was the question! – Dr. Sonnhard Graubner Nov 27 '16 at 14:26

$$-\frac a5\left(2\times \frac {a^2}5\right)^2=-\frac a5\times \frac {4a^4}{25}=-\frac {4a^5}{125}$$

$$5\left(-\frac a5\right)^3\left(2\times \frac {a^2}5\right)=-\frac {a^3}{25}\left(\frac {2a^2}5\right)=-\frac {2a^5}{125}$$ And $$-a^2\left(-\frac a5\right)\left(2\times \frac {a^2}{5}\right)=\frac {a^3}5\left(2\times\frac {a^2}5\right)=\frac {2a^5}{25}$$

Combining, we get this:$$\color{blue}{-\frac a5\left(\frac {2a^2}5\right)^2}+\color{red}{5\left(-\frac a5\right)^3\left(\frac {2a^2}5\right)}\color{brown}{-a^2\left(-\frac a5\right)\left(\frac {2a^2}5\right)}\\ =\color{blue}{-\frac {4a^5}{125}}\color{red}{-\frac {2a^5}{125}}\color{brown}{+\frac {2a^5}{25}}=-\frac {6a^5}{125}+\frac {2a^5}{25}=\boxed{\frac {4a^5}{125}}$$

For where you went wrong, you put $\frac {-4a^5}{25}$ instead of $-\frac {4a^5}{125}$. Most likely, you forgot to square the denominator of the fraction in $\left(\frac {2a^2}5\right)^2$.

• What question are you answering? – GFauxPas Nov 27 '16 at 14:54
• @GFauxPas The top half, I'm evaluating the problem. The bottom half, I'm answering the OP's question on What have I done wrong? – Frank Nov 27 '16 at 14:55

You start with an equation but I suppose you are only concerned about the left hand side of the equation.

$$-\frac{a}{5}\left(2\times\frac{a^2}{5}\right)^2+5\left(-\frac{a}{5}\right)^3\left(2\times\frac{a^2}{5}\right) -a^2\left(-\frac{a}{5}\right)\left(2\times\frac{a^2}{5}\right)\tag{1}$$ I get $$\frac{-4a^5}{25}-\frac{2a^5}{125}+\frac{2a^5}{25}\tag{2}$$ (...) $$-\frac{12a^5}{125}\tag{3}$$

How to find the error?

Plug in some numbers for the variablee $a$ and compare, e.g $a=5$ in $(1)$ will give

$$-\frac{5}{5}\left(2\times\frac{5^2}{5}\right)^2+5\left(-\frac{5}{5}\right)^3\left(2\times\frac{5^2}{5}\right) -5^2\left(-\frac{5}{5}\right)\left(2\times\frac{5^2}{5}\right) \\ =-1(10)^2+5(-1)^3(10)-25(-1)(10)\\ =-100-50+250\\=100$$ but $(2)$ will give $$\frac{-4\cdot 5^5}{25}-\frac{2\cdot 5^5}{125}+\frac{2\cdot 5^5}{25}\\ =-4\cdot 5^3-2\cdot5^2+2\cdot5^3\\ =-500-50+250\\ =300$$ So you introduced an error when you calculated $(2)$ from $(1)$

You can plug in arbitrary numbers for $a$ and use a calculator to verify/disprove your calculation. If a number you plug in verifies your calculation that does not mean that there is no error in the calculation, maybe other values will disprove the calculation. But if a number disproves a calculation you can be sure that there is an error in your calculation and try to find it by using the same method for the substeps of your calculation.

So next you check each term in this step. Ifyou plug in $a=6$ in the first term
$$-\frac{a}{5}\left(2\times\frac{a^2}{5}\right)^2$$ and in the first term of your your result $$\frac{-4a^5}{25}$$

and use a calculator you will get something like $-248.832$ and $-1244.16$ (You almost always have a calculator. I simply put -6/5*(2*6^2/5)^2 and -4*6^5/25 into Google to get these results). So there is an error in transforming this term.

You can often use such a technique to find errors in your calculations.

If you have access to a CAS like Mathematica you can use this tool to check your intermediate results.