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

**normalguy** · June 16th 2011, 06:52 AM

Proof that: $sqrt(ab-ac)=asqrt(b-c)$
without the use of substitution

Thx

**Prove It** · June 16th 2011, 06:56 AM

It doesn't.

$\displaystyle \sqrt{ab - ac} = \sqrt{a(b - c)} = \sqrt{a}\sqrt{b - c}$ , not $\displaystyle a\sqrt{b-c}$ .

**HallsofIvy** · June 16th 2011, 07:15 AM

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.)

**Prove It** · June 16th 2011, 07:41 AM

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 $\displaystyle \sqrt{ab - ac} = \sqrt{a}\sqrt{b-c}$ ...

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