Proof that: $\displaystyle sqrt(ab-ac)=asqrt(b-c)$

without the use of substitution

Thx

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- Jun 16th 2011, 06:52 AMnormalguyproof that sqrt(ab-ac)=a sqrt(b-c)
Proof that: $\displaystyle sqrt(ab-ac)=asqrt(b-c)$

without the use of substitution

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

$\displaystyle \displaystyle \sqrt{ab - ac} = \sqrt{a(b - c)} = \sqrt{a}\sqrt{b - c}$, not $\displaystyle \displaystyle a\sqrt{b-c}$. - Jun 16th 2011, 07:15 AMHallsofIvyRe: 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 $\displaystyle \sqrt{a^2b- a^2c}= a\sqrt{b- c}$?

(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

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

($\displaystyle \sqrt{a^2}= a$ if a is non-negative. $\displaystyle \sqrt{a^2}= -a$ if a is negative.) - Jun 16th 2011, 07:41 AMProve ItRe: proof that sqrt(ab-ac)=a sqrt(b-c)