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The other day during a seminar, in a calculation, a fellow student encountered this expression: $$\sqrt{3^2 + (6t)^2 + (6t^2)^2}$$ He, without much thinking, immediately wrote down: $$(6t^2+3)$$

What bothers me, is that I didn't see that. Although I suspected there might be some binomial expansion, I was too confused by the middle term to see it. Even after actually seeing the answer, it took me a good deal to realize what's happening. My initial idea, by the way, was to substitute for $t^2$ and solve a quadratic equation. Horribly slow.

My questions is perhaps tackling a greater problem. I now see what happened, I can "fully" understand binomial expansions, yet I feel like I might end up encountering a similar expression and yet again not be able to see the result. My questions is thus:

How can I properly practice so I don't repeat this mistake?

This is perhaps too much of a soft-question, but I feel like many others have been in a position of understanding a problem, yet not feeling like they would be able to replicate it, and might actually have an answer that goes well beyond the obvious "just practice more".

Thanks for any input.

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5 Answers 5

up vote 2 down vote accepted

This may actually be a matter of "just practice more." If you've done enough factoring, you'll recognize the coefficients $4$, $4$, and $1$ from $(x+2)^2 = x^2 + 4x + 4$. Taking out the $3$ is also a matter of practicing factorization - noticing that the terms all have a common factor.

Being able to do this mentally is, yet again, a matter of practice - this time of mental arithmetic. It may not be that your fellow student consciously set aside time at some point in his life toward practicing mental arithmetic.

More likely, he has experience doing things like that in math classes - when teachers write expressions on the board, he immediately starts thinking about solutions in his head ...

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After a bit of thinking I decided to choose this as the accepted answer, as it is the most direct answer to my question. Thanks all the others for letting me see other aspects / approaches (regardless of only subtle differences). –  Dahn Jahn Nov 29 '13 at 23:16

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $$ \root{3^{2} + \bracks{6t}^{2} + \bracks{6t^{2}}^2} = 3\root{4t^{4} + 4t^{2} + 1} = 3\root{\pars{2t^{2} + 1}^{2}} = 3\pars{2t^{2} + 1} = 6t^{2} + 3 $$

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This is the way I would have done it. –  Lubin Nov 29 '13 at 1:44
    
@Lubin It's straightforward. –  Felix Marin Nov 29 '13 at 14:58
    
I suppose it is, yet it gives me valuable insight into how other people see this. Thank you. –  Dahn Jahn Nov 29 '13 at 23:14
    
@DahnJahn You're welcome. Thanks. –  Felix Marin Nov 29 '13 at 23:15

$\bigcirc^2+\triangle+\square^2$ is a complete square if $\left|\triangle\right|=2\cdot\bigcirc\cdot\triangle$.

For example, $49x^2-42x+9$ is a complete square since the only candidate for $\triangle$ in $42x$, and indeed $42x=2\cdot7x\cdot3$. In our special case we have three candidates for $\triangle$; your fellow student had three candidates for $\triangle$. “Without much thinking” -- a bit of thinking was implied, since the three choices -- he verified that $(6t)^2=2\cdot3\cdot6t^2$.

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All you need to see is

$$6^2=2 \cdot 3 \cdot 6$$

Then $$\sqrt{3^2 + (6t)^2 + (6t^2)^2}=\sqrt{3^2 + 2 \cdot 3 \cdot (6t^2) + (6t^2)^2}$$

Anyhow, in general an expression like this is a monomial times a perfect square if and only if the product of the largest and smallest degree is exactly the square of half of the middle term. In this case, the monomial is exactly the smallest term in the expression.

In this case, a fast check if this is a perfect square would be to see that $3^2$ is a perfect square and :

$$3^2 \cdot (6t^2)^2 \stackrel{?}{=} \left( \frac{(6t)^2}{2}\right)^2$$

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How do I write an $=$ with $?$ over it? –  N. S. Nov 28 '13 at 23:11
    
\stackrel{?}{=} –  Dahn Jahn Nov 28 '13 at 23:15
    
@DahnJahn Tyvm. –  N. S. Nov 28 '13 at 23:21

The radicand is obviously a multiple of $9$. It factors to $9(1+(2t)^2+(2t^2)^2)$, and that simplifies to $9(1+4t^2+4t^4)$. It is usual to write polynomials in decreasing order of powers, so you get $9(4t^4+4t^2+1)$, which equals $9(4(t^2)^2+4t^2+1)$. Someone with experience in factoring things may recognize the last expression as $9(2t^2+1)^2$ (you have something of the form "$4x^2+4x+1$"). When you take the square root of that, you get $\sqrt{9}\sqrt{(2t^2+1)^2}=3(2t^2+1)=6t^2+3$.

This is not the fastest way to do it. It just takes practice. I think it is helpful here to factor out the $9$ to simplify and recognize the radicand as a quadratic in $t^2$, so you are really factoring a quadratic.

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