## [MOVED] roots and the least common denominator

Simplificatation, evaluation, linear equations, linear graphs, linear inequalities, basic word problems, etc.
Tennise
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Joined: Sat Jun 06, 2009 7:12 am
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### [MOVED] roots and the least common denominator

To find the least common denominator of ${2 \over 3a}+{4 \over a^2}$ the book I'm reading says to multiply them to get a common denominator of $3a^2: {2a \over 3a^2}+{12 \over 3a^2} = {2a+12 \over 3a^2}$
But wouldn't it be more efficient to find the square roots of the second fraction before multiplying? ${2 \over 3a}+sqrt{ 4\over a^2} = {2 \over 3a}+{2 \over a}\cdot{3 \over 3} = {2 \over 3a} + {6 \over 3a} = {8 \over 3a}$

jaybird0827
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Location: NC
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### Re: roots and the least common denominator

To find the least common denominator of ${2 \over 3a}+{4 \over a^2}$ the book I'm reading says to multiply them to get a common denominator of $3a^2: {2a \over 3a^2}+{12 \over 3a^2} = {2a+12 \over 3a^2}$
But wouldn't it be more efficient to find the square roots of the second fraction before multiplying? ${2 \over 3a}+sqrt{ 4\over a^2} = {2 \over 3a}+{2 \over a}\cdot{3 \over 3} = {2 \over 3a} + {6 \over 3a} = {8 \over 3a}$
Where do you get the idea that the square root of a fraction is equal to the fraction?

Please compare the value of $\frac{4}{a^2}$ with $\frac{2}{a}$ by plugging a random value such as $7$ into each of these expressions for $a$. Are you sure that $\frac{4}{a^2}\,=\,\frac{2}{a}$ for all $a$?

Also, you may find it helpful to review lcm and gcf here: http://www.purplemath.com/modules/lcm_gcf.htm

Tennise
Posts: 7
Joined: Sat Jun 06, 2009 7:12 am
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### Re: roots and the least common denominator

Oh, I see now... Thought I could cheat my way through and remove the exponent by squaring, guess not!

Thanks for your help, and the link, although my problem wasn't LCM/GCF here :P

Edit: I just noticed you moved this thread, sorry for posting in the wrong area, arithmetic's description listed fractions...