Square root problem

**sarahh** · December 21st 2010, 02:48 AM

If $\sqrt{a + b} = \sqrt{a} + \sqrt{b}$ , which of the following must be true?

A. $a = b$
B. $a = 0$ and $b = 0$
C. $a = 0$ or $b = 0$
D. $a + b > 1$
E. $a = 1$ and $b > 0$

I mean clearly C works but why don't the other choices work?

**HallsofIvy** · December 21st 2010, 03:17 AM

Originally Posted by sarahh

If $\sqrt{a + b} = \sqrt{a} + \sqrt{b}$ , which of the following must be true?

A. $a = b$
B. $a = 0$ and $b = 0$
C. $a = 0$ or $b = 0$
D. $a + b > 1$
E. $a = 1$ and $b > 0$

I mean clearly C works but why don't the other choices work?

For each of A, B, D, E, let a= 1, b= 0

**sarahh** · December 21st 2010, 03:33 AM

Hmm so wait Ivy--would you say that for this question all the answers are correct but choice C is more correct?? (Because of your values)

**BAdhi** · December 21st 2010, 04:01 AM

no he says only C is correct :-)

**Prove It** · December 21st 2010, 04:03 AM

Originally Posted by sarahh

If $\sqrt{a + b} = \sqrt{a} + \sqrt{b}$ , which of the following must be true?

A. $a = b$
B. $a = 0$ and $b = 0$
C. $a = 0$ or $b = 0$
D. $a + b > 1$
E. $a = 1$ and $b > 0$

I mean clearly C works but why don't the other choices work?

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

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

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

$\displaystyle 0 = 2\sqrt{a}\sqrt{b}$ .

For this to be true, $\displaystyle a = 0, b = 0$ or $\displaystyle a=b=0$ .

The one that must be true is C.

**sarahh** · December 21st 2010, 04:09 AM

Thanks Prove It! That's quite ingenius and beautiful actually

**Archie Meade** · December 21st 2010, 02:39 PM

Originally Posted by sarahh

If $\sqrt{a + b} = \sqrt{a} + \sqrt{b}$ , which of the following must be true?

A. $a = b$

B. $a = 0$ and $b = 0$

C. $a = 0$ or $b = 0$

D. $a + b > 1$

E. $a = 1$ and $b > 0$

Or...

$\sqrt{a+b}=\sqrt{a}+\sqrt{b}\Rightarrow\sqrt{a+b}\ left[\sqrt{a}-\sqrt{b}\right]=\left[\sqrt{a}+\sqrt{b}\right]\left[\sqrt{a}-\sqrt{b}\right]$

$\Rightarrow\sqrt{a+b}\sqrt{a}-\sqrt{a+b}\sqrt{b}=a-\sqrt{a}\sqrt{b}+\sqrt{a}\sqrt{b}-b$

$\Rightarrow\sqrt{a^2+ab}-\sqrt{ab+b^2}=a-b\Rightarrow\ ab=0$

I mean clearly C works but why don't the other choices work?

A.

$a=b$

$\sqrt{2a}=\sqrt{2}\sqrt{a}$

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

Not equal.

B.

$\sqrt{0+0}=\sqrt{0}+\sqrt{0}$

however the equation is true for either one of the two values being non-zero.

Hence it is unnecessary that both be zero.

D.

$a+b>1$

$a=1,\;\;\ b=0\Rightarrow\sqrt{1+0}=\sqrt{1}+\sqrt{0}$

It's unnecessary for $a+b>1$

E.

$a=1,\;\;b=0$ works, hence the condition is not required.

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