Hard? problem solving

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Hard? problem solving

by cutty61 » Sat Apr 09, 2011 10:37 am
(√(9+√80) ) (√(9-√80) )^2= ?





I get 18. Answer is credited as 20.

Where am I going wrong?

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by Ian Stewart » Sat Apr 09, 2011 12:15 pm
cutty61 wrote:(√(9+√80) ) (√(9-√80) )^2= ?

I get 18. Answer is credited as 20.

Where am I going wrong?
I think you're missing a '+' sign in the question; it should read:

[ sqrt(9 + sqrt(80)) + sqrt(9 - sqrt(80)) ]^2 = ?


This certainly is not an easy question, because it tests several rules, and it's difficult to avoid a somewhat lengthy calculation. It is really just a complicated way to test your facility with the basic factoring patterns and square root rules. To start, we can use the pattern:

(x + y)^2 = x^2 + y^2 + 2xy

If we expand the expression using the pattern above, we get:

= [sqrt(9 + sqrt(80))]^2 + [sqrt(9 - sqrt(80))]^2 + 2*[sqrt(9 + sqrt(80))]*[sqrt(9 - sqrt(80))]


We now can use the fact that [sqrt(x)]^2 is just equal to x to simplify the first and second of the three expressions in our sum. For the third expression, we can use the fact that (sqrt(x))(sqrt(y)) = sqrt(xy). We'll then be multiplying two terms which are in a difference of squares pattern -- that is, our product will look like (a+b)(a-b), so will be equal to a^2 - b^2:

= 9 + sqrt(80) + 9 - sqrt(80) + 2*sqrt[(9 + sqrt(80))(9 - sqrt(80))]

= 18 + 2*sqrt( 9^2 - (sqrt(80))^2)

= 18 + 2*sqrt(81 - 80)

= 18 + 2*sqrt(1)

= 18 + 2

= 20

So in the end it all simplifies nicely, though it starts out looking like a mess. You can estimate rather quickly with this question to see that the answer should be something close to 18, and as I recall, there are only two answer choices which are at all reasonable here, 18 and 20, so an estimate at least gets you to a 50-50 guess. If you do that estimate first, then you can stop working as soon as you realize the answer will be greater than 18, though that really doesn't save much time - the hard steps are the first ones. To actually arrive at the answer, I don't think there's any way to avoid the steps above, or some equivalent calculation.
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