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If a – b = 16 and \(\sqrt{a} + \sqrt{b} = 8\) , what is the value \(\sqrt{ab}\) ?

(A) 2 (B) 4 (C) 8 (D) 12 (E) 15

IMPORTANT: (√a + √b)(√a - √b) = a - b [this is a cute version of difference of squares] We're told that a - b = 16 AND √a + √b = 8 Plug these values into the equation to get: (8)(√a - √b) = 16 From this, we can see that √a - √b = 2

We have: √a + √b = 8 √a - √b = 2

When we ADD the two equations, we get: 2√a = 10, which means √a = 5 When we SUBTRACT the bottom equation from the top, we get: 2√b = 6, which means √b = 3

So, √(ab) = (√a)(√b) = (5)(3) = 15

Answer: E

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Brent Hanneson – Creator of greenlighttestprep.com Sign up for my free GRE Question of the Dayemails

Re: If a – b = 16 and [#permalink]
23 Nov 2017, 22:59

2

This post received KUDOS

I am posting my answer to exemplify how important it is to make sure you are taking the shortest possible route when solving a problem. I didn't do that and look at the monster I got when I tried to use substitution:

A - B = 16 A = 16 + B

Now I substituted the A for 16 + B in the other equation:

√A + √B) = 8 √(16 + B) + √B = 8

√(16 + B) = 8 - √B

Both sides are squared trying to get rid of the square root:

(√(16 + B))^2 = (8 - √B)^2 16 + B = (8 - √B) x (8 - √B) 16 + B = 64 -16√B + B -64 -B -64 -B -48 = -16√B

Square both sides again to get rid of the radical:

(-48)^2 = (-16√B)^2 2304 = 256B B = 2304/256 B = 9

Now substitute the B in the first equation:

A - B = 16 A - 9 = 16 A = 25

Now plug in A and B in the third equation:

√AB = ? √(9) x (25) √225 = 15

Sorry it is a little confusing. This is my first post. I am a little pround I got the final result after this mathmatical ordeal.

Re: If a – b = 16 and [#permalink]
17 Jul 2019, 10:15

1

This post received KUDOS

Expert's post

mibad wrote:

roota+rootb = 8 (roota+rootb)^2=8^2 a + 2(√a)(√b) + b = 64 2√(ab) = 64 - a - b

a - b = 16

2√(ab) = 64 - 16 √(ab) = 48/2 √(ab) = 24

I've highlighted a problem above. In short, 64 - a - b is NOT the same as 64 - (a - b)

You're correct to say that: 2√(ab) = 64 - a - b We can rewrite the right side as: 2√(ab) = 64 - (a + b) The problem is that we don't know the value of a+b.

Does that help?

Cheers, Brent
_________________

Brent Hanneson – Creator of greenlighttestprep.com Sign up for my free GRE Question of the Dayemails

Re: If a – b = 16 and [#permalink]
02 May 2020, 03:31

1

This post received KUDOS

I don't know if I was extremely lucky or it was intuition.

I looked at root(a) + root(b) = 8 and the a - b = 16 equations, and also the options. All are integers, no points or fractions or roots in the values. So I figured a and b have to be integers.

root(a) + root(b) = 8 This means that the roots of a and b are less than 8 because roots cannot be negative.

The first option that popped in my mind was 5 + 3 = 8. So I plugged them in the other equation and found that indeed, 25 - 9 = 16.

So root(ab) = root(a)*root(b) = 3*5 = 15.

greprepclubot

Re: If a – b = 16 and
[#permalink]
02 May 2020, 03:31