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by rijul007 » Wed Dec 14, 2011 5:03 pm
winniethepooh wrote:√{3√80 + 3/(9+4√5)} = ?
√{3√{4*4*5} + 3(9-4√5)/(9-4√5)(9+4√5)}
√{12√5 + 3(9-4√5)/81-80?
√{12√5 + 27 - 12√5}
√27
3√3
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by GmatMathPro » Wed Dec 14, 2011 5:03 pm
winniethepooh wrote:√{3√80 + 3/(9+4√5)} = ?
First, focus on the expression 3/(9+4√5). To simplify this, multiply the top and bottom by the conjugate of the bottom. The conjugate of an expression like this is just the same expression with the opposite sign in the middle. So in this case, it would be 9-4√5.

[3/(9+4√5)][(9-4√5)/(9-4√5)] = (27-12√5)/(81-36√5+36√5-80) = (27-12√5)/1 = 27-12√5. Plug this in to the original expression to get:

√{3√80+27-12√5}. Notice that 3√80 = 3√(16*5) = 12√5, so the entire expression simplifies to √27, which simplifies to [spoiler]3√3[/spoiler]
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by Anurag@Gurome » Wed Dec 14, 2011 8:47 pm
winniethepooh wrote:√{3√80 + 3/(9+4√5)} = ?
√{3√80 + 3/(9+4√5)} = √{3 * 4√5 + 3/(9 + 4√5)} = √{12√5 + 3/(9 + 4√5)}
Now, 3/(9 + 4√5) = 3(9 - 4√5)/(9 + 4√5)(9 - 4√5) = 3(9 - 4√5)/(81 - 80) = 3(9 - 4√5)= 27 - 12√5
Therefore, √{3√80 + 3/(9+4√5)} = √{12√5 + 27 - 12√5} = √27 = [spoiler]3√3[/spoiler]
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by GMATGuruNY » Thu Dec 15, 2011 3:57 am
winniethepooh wrote:√{3√80 + 3/(9+4√5)} = ?
If the answer choices are sufficiently different, we can ballpark the answer quickly:

√(3√80 + 3/(9+4√5))

≈ √( 3*√81 + smaller number/bigger number )

= √(3*9 + fraction)

= √(27 + fraction).

The fraction will have little effect on the result.
Thus, we can ignore it in our calculations:

√27 ≈ 3√3.
Last edited by GMATGuruNY on Thu Dec 15, 2011 6:53 pm, edited 2 times in total.
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by winniethepooh » Thu Dec 15, 2011 6:18 pm
is there no way other than conjugating that this problem can be solved?
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by winniethepooh » Thu Dec 15, 2011 6:22 pm
GMATGuruNY wrote:
winniethepooh wrote:√{3√80 + 3/(9+4√5)} = ?
If the answer choices are sufficiently different, we can ballpark the answer quickly:

√(3√80 + 3/(9+4√5))

≈ √(3*9 + fraction)

≈ √27

≈ 3√7.
Hi Mitch, I didn't understand how did you get to the step in bold red above.
Just to correct √27 = 3√3
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by rijul007 » Thu Dec 15, 2011 6:37 pm
winniethepooh wrote:
GMATGuruNY wrote:
winniethepooh wrote:√{3√80 + 3/(9+4√5)} = ?
If the answer choices are sufficiently different, we can ballpark the answer quickly:

√(3√80 + 3/(9+4√5)

≈ √(3*9 + fraction)

≈ √27

≈ 3√7.
Hi Mitch, I didn't understand how did you get to the step in bold red above.
Just to correct √27 = 3√3
√(3√80 + 3/(9+4√5))


to make the calculation much easier we replace √80 with √81 as they are approximately the same

√(3√81 + 3/(9+4√5))
√(3*9 + 3/(9+4√5))


4√5 = 2.23 * 4 = 9 approx.

3/(9+4√5) ≈ 3/(9+9) ≈ 1/6 ≈ 0.17
this is almost zero, so he neglected it

so the expression would be √(3*9 + 0.17) ≈ √3*9 ≈ 3√3
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by GMATGuruNY » Thu Dec 15, 2011 6:56 pm
winniethepooh wrote:
GMATGuruNY wrote:
winniethepooh wrote:√{3√80 + 3/(9+4√5)} = ?
If the answer choices are sufficiently different, we can ballpark the answer quickly:

√(3√80 + 3/(9+4√5))

≈ √(3*9 + fraction)

≈ √27

≈ 3√7.
Hi Mitch, I didn't understand how did you get to the step in bold red above.
Just to correct √27 = 3√3
Please revisit my post above, in which I fleshed out the explanation.
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by winniethepooh » Fri Dec 16, 2011 4:29 am
Wow Mitch, as always your explanations make such a lot of sense!!

That's really helpful in saving time!!

Thanks a lot.
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