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2nd degree equations

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2nd degree equations

by money9111 » Sun Feb 14, 2010 7:03 pm
A positive number X is multiplied by 2, and this product is then divided by 3. If the positive square root of the results of these two operations equals x, what is the value of x?

A. 9/4
B. 3/2
C. 4/3
D. 2/3
E. 1/2

OA D

In solving this equation I got to 2x=3(x^2), but then for some reason decided to divide by 2 (which is clearly wrong because then I got stuck)... my question is... how do you know to then subtract 2x from both sides, and then solve for X?
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by ejose » Sun Feb 14, 2010 7:50 pm
Money9111 --

You may be over-thinking this one :)

In order to isolate x here, ultimately you'll have to divide. I suppose you could factor it, too, but simple division is going to yield a result faster.

To start, the expression says ...

1. A positive number X is multiplied by 2, and this product is then divided by 3 :: 2/3x
2. .... If the positive square root of the results of these two operations equals x :: sqrt(2/3x)=x

then ..

3. square both sides :: 2/3x = x^2
4. divide both sides by x :: 2/3x / x = x^2 / x == 2/3 = x

and that's your answer.

I think that what might be tripping you up is that you're evaluating the equation as "2x over three" instead of "x times 2/3" .. which is making you want to get rid of that three. I could be wrong :)

-Eric
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by money9111 » Sun Feb 14, 2010 8:07 pm
thank you for that explanation... i was probably overthinking it... didn't think to just simply divide both sides by x... i know now though!
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