If a and b are positive numbers, explain how the value of

(sqrt{a}*sqrt{b})^2 compares with the value of a + b.

I used FOIL and ended up with a + 2(sqrt{ab}) + b, which does not equal a + b.

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- May 5th 2009, 09:12 PM #1

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- May 5th 2009, 09:30 PM #2

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- May 5th 2009, 09:53 PM #3
$\displaystyle (\sqrt{a}\sqrt{b})^2 = ab$.

Your statement will only be true if

$\displaystyle ab = a + b$

$\displaystyle 0 = a - ab + b$

$\displaystyle 0 = a(1 - b) + b$

$\displaystyle -b = a(1 - b)$

$\displaystyle -\frac{b}{1 - b} = a$

$\displaystyle \frac{b}{b - 1} = a$.

So your situation will only be true for positive values of $\displaystyle a, b$ so that $\displaystyle a = \frac{b}{b - 1}$.

- May 6th 2009, 05:12 AM #4

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- May 6th 2009, 02:36 PM #5
$\displaystyle (\sqrt{a} + \sqrt{b})^2 = a + 2\sqrt{a}\sqrt{b} + b$

$\displaystyle = a + 2\sqrt{ab} + b$.

The statement will be true if

$\displaystyle a + 2\sqrt{ab} + b = a + b$

$\displaystyle 2\sqrt{ab} = 0$

$\displaystyle \sqrt{ab} = 0$

$\displaystyle ab = 0$.

This will only happen if one or both of a and b is 0.

So your equation only holds true for if one or both of a and b is 0.

- May 6th 2009, 04:21 PM #6

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You used "foil" on what? Of course, a+ 2sqrt{ab}+ b is not equal to a+ b, why should it be? Were you asked to about (sqrt{a}*sqrt{b})^2 or (sqrt{a}+ sqrt{a})^2? And you weren't asked to show these were equal, only to "compare" them. Which is larger, (sqrt{a}+ sqrt{b})^2 or (a+ b)^2?

- May 6th 2009, 06:12 PM #7

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