# proof that sqrt(ab-ac)=a sqrt(b-c)

• Jun 16th 2011, 06:52 AM
normalguy
proof that sqrt(ab-ac)=a sqrt(b-c)
Proof that: $sqrt(ab-ac)=asqrt(b-c)$
without the use of substitution

Thx
• Jun 16th 2011, 06:56 AM
Prove It
Re: proof that sqrt(ab-ac)=a sqrt(b-c)
It doesn't.

$\displaystyle \sqrt{ab - ac} = \sqrt{a(b - c)} = \sqrt{a}\sqrt{b - c}$, not $\displaystyle a\sqrt{b-c}$.
• Jun 16th 2011, 07:15 AM
HallsofIvy
Re: proof that sqrt(ab-ac)=a sqrt(b-c)
As Prove It said, you can't prove that- it's not true.

Did you mean $\sqrt{a^2b- a^2c}= a\sqrt{b- c}$?
(assuming that a is non-negative)

I'm not sure what you mean by "without the use of substitution"- I see nothing to substitute there- but
$\sqrt{a^2b- a^2c}= \sqrt{a^2(b- c)}= \sqrt{a^2}\sqrt{b- c}= a\sqrt{b- c}$

( $\sqrt{a^2}= a$ if a is non-negative. $\sqrt{a^2}= -a$ if a is negative.)
• Jun 16th 2011, 07:41 AM
Prove It
Re: proof that sqrt(ab-ac)=a sqrt(b-c)
Quote:

Originally Posted by HallsofIvy
As Prove It said, you can't prove that- it's not true.

Did you mean $\sqrt{a^2b- a^2c}= a\sqrt{b- c}$?
(assuming that a is non-negative)

I'm not sure what you mean by "without the use of substitution"- I see nothing to substitute there- but
$\sqrt{a^2b- a^2c}= \sqrt{a^2(b- c)}= \sqrt{a^2}\sqrt{b- c}= a\sqrt{b- c}$

( $\sqrt{a^2}= a$ if a is non-negative. $\sqrt{a^2}= -a$ if a is negative.)

Alternatively, maybe the OP was being asked to prove $\displaystyle \sqrt{ab - ac} = \sqrt{a}\sqrt{b-c}$...