# Math Help - a and b

1. ## a and b

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.

2. And you would be 100% correct

3. Originally Posted by magentarita
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.
$(\sqrt{a}\sqrt{b})^2 = ab$.

Your statement will only be true if

$ab = a + b$

$0 = a - ab + b$

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

$-b = a(1 - b)$

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

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

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

4. ## sorry...but

Originally Posted by Prove It
$(\sqrt{a}\sqrt{b})^2 = ab$.

Your statement will only be true if

$ab = a + b$

$0 = a - ab + b$

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

$-b = a(1 - b)$

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

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

So your situation will only be true for positive values of $a, b$ so that $a = \frac{b}{b - 1}$.
I am sorry. I meant to type (sqrt{a} + sqrt{b})^2 in the original question. Can you show me now?

5. $(\sqrt{a} + \sqrt{b})^2 = a + 2\sqrt{a}\sqrt{b} + b$

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

The statement will be true if

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

$2\sqrt{ab} = 0$

$\sqrt{ab} = 0$

$ab = 0$.

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.

6. Originally Posted by magentarita
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.
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?

7. ## the book

Originally Posted by Prove It
$(\sqrt{a} + \sqrt{b})^2 = a + 2\sqrt{a}\sqrt{b} + b$

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

The statement will be true if

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

$2\sqrt{ab} = 0$

$\sqrt{ab} = 0$

$ab = 0$.

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.
The question is asking to compare. After using FOIL, I saw that a + 2(sqrt{ab}) + b is > than a + b, which is the book's answer.