for kids - 3

logistic_guy

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\(\displaystyle \textcolor{indigo}{\bold{Solve.}}\)

\(\displaystyle |4n - 15| = |n|\)
 
[imath](4n - 15)^2 = n^2[/imath]
😍😍
You must be a child prodigy or something, Smith. I have never thought to solve it in this way!

\(\displaystyle \textcolor{purple}{\bold{Smith's \ method.}}\)


[imath]|4n - 15| = |n|[/imath]

[imath](4n - 15)^2 = n^2[/imath]

[imath]15n^2 - 120n + 225 = 0[/imath]

[imath](5n - 15)(3n - 15) = 0[/imath]

This gives:

\(\displaystyle n = \textcolor{blue}{\bold{3}}\)
Or
\(\displaystyle n = \textcolor{red}{\bold{5}}\)
 
Don't worry Agent. In my eyes, you're an adult.

\(\displaystyle \textcolor{darkgreen}{\bold{My \ method.}}\)


[imath]|4n - 15| = |n|[/imath]

Get rid of the absolute value and use all combination of signs. In other words, I have to solve the following:

\(\displaystyle \textcolor{red}{\bold{1.}} \ 4n - 15 = n\)
\(\displaystyle \textcolor{blue}{\bold{2.}} \ -(4n - 15) = -n\)
\(\displaystyle \textcolor{indigo}{\bold{3.}} \ -(4n - 15) = n\)
\(\displaystyle \textcolor{grey}{\bold{4.}} \ (4n - 15) = -n\)


Because of symmetry, I have to solve only:



\(\displaystyle \textcolor{red}{\bold{1.}} \ 4n - 15 = n\)
\(\displaystyle \textcolor{grey}{\bold{4.}} \ (4n - 15) = -n\)

Solving \(\displaystyle \textcolor{red}{\bold{One}}\)

\(\displaystyle 4n - 15 = n\)
\(\displaystyle 4n - n = 15\)
\(\displaystyle 3n = 15\)
\(\displaystyle n = \frac{15}{3} = \textcolor{red}{\bold{5}}\)



Solving \(\displaystyle \textcolor{grey}{\bold{Four}}\)

\(\displaystyle (4n - 15) = -n\)
\(\displaystyle 4n + n = 15\)

\(\displaystyle 5n = 15\)

\(\displaystyle n = \frac{15}{5} = \textcolor{grey}{\bold{3}}\)

Therefore, the difficulty of your method is that: Can a \(\displaystyle 3\) years old kid factor this?

[imath]\textcolor{indigo}{15n^2 - 120n + 225 = 0} \longrightarrow \textcolor{blue}{(5n - 15)(3n - 15) = 0}[/imath]

😉
 
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