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May 12th, 2015, 06:25 AM   #1
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rationalize a denominator last task

Task 1.25 C) This is the task

and this is the answer in the book

I do understand that (x squareroot x - y square root y) ( squareroot x - squaretoot y/(squareroot x)^2 - (squareroot y)^2 is x -y Also will be (x square root x + y square root y)^2 + (square root x - squareoot y)^2/ x -y I also understand that you use square sentence three in this example.

But i don't know what happens after that and why. I am stuck in that part. Would be fine if someone could help me.
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May 12th, 2015, 07:26 AM   #2
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Maybe it will be easier to consider the denominator and numerator separately and show exactly what happens step by step.

So, after multiplying the top and bottom by the conjugate of the denominator (i.e. $\displaystyle \sqrt{x} + \sqrt{y}$)

Denominator:

Start with:
$\displaystyle (\sqrt{x} - \sqrt{y})(\sqrt{x} + \sqrt{y})$

Multiply out brackets:
$\displaystyle =\sqrt{x}\sqrt{x} + \sqrt{y}\sqrt{x} - \sqrt{x}\sqrt{y} - \sqrt{y}\sqrt{y}$

The two middle terms subtract with each other.
$\displaystyle =\sqrt{x}\sqrt{x} - \sqrt{y}\sqrt{y}$

Because $\displaystyle \sqrt{a}\sqrt{a} = \sqrt{a \times a} = \sqrt{a^2} = a$, the next step is to apply this to the x-part and the y-part:
$\displaystyle =x -y$

Now for the numerator:

Start with:
$\displaystyle (x\sqrt{x} + y\sqrt{y})(\sqrt{x} + \sqrt{y})$

Multiply out the brackets:
$\displaystyle =x\sqrt{x}\sqrt{x} + x\sqrt{y}\sqrt{x} + y\sqrt{x}\sqrt{y} - y\sqrt{y}\sqrt{y}$

Because $\displaystyle \sqrt{a}\sqrt{a} = \sqrt{a \times a} = \sqrt{a^2} = a$, the next step is to apply this to the first and last terms:
$\displaystyle =x\times x + x\sqrt{y}\sqrt{x} + y\sqrt{x}\sqrt{y} + y \times y$

Use index notation (i.e. $\displaystyle a \times a = a^2$):
$\displaystyle =x^2 + x\sqrt{y}\sqrt{x} + y\sqrt{x}\sqrt{y} + y^2$

Factorise $\displaystyle \sqrt{x}\sqrt{y}$ from the middle two terms:
$\displaystyle =x^2 + \sqrt{y}\sqrt{x}(x+y) + y^2$

Rearrange order slightly:
$\displaystyle =x^2 + y^2 + (x+y)\sqrt{y}\sqrt{x}$

Putting together, the final expression is:

$\displaystyle \frac{x^2 + y^2 + (x+y)\sqrt{y}\sqrt{x}}{x-y}$
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May 12th, 2015, 03:55 PM   #3
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Because $\displaystyle \sqrt{a}\sqrt{a} = \sqrt{a \times a} = \sqrt{a^2} = a$
That's not true.

$\displaystyle \sqrt{a}\sqrt{a} \ \ne \ \sqrt{a \times a} $


$\displaystyle \sqrt{a \times a} \ = \ \sqrt{a^2} \ = \ |a|$


It's because $\displaystyle \ \ \sqrt{a}\sqrt{a} \ = \ (\sqrt{a})^2 \ = \ a.$
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May 12th, 2015, 05:15 PM   #4
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That's not true.
$\displaystyle \sqrt{a}\sqrt{a} \ \ne \ \sqrt{a \times a} $
I don't understand. Could you provide a positive value of "a" where the two sides are not equivalent?
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May 13th, 2015, 11:26 AM   #5
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Quote:
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I don't understand. Could you provide a positive value of "a" where the two sides are not equivalent?
Quote:
Originally Posted by Math Message Board tutor View Post
That's not true.

$\displaystyle \sqrt{a}\sqrt{a} \ \ne \ \sqrt{a \times a} $
No, it's that the two sides are not equal in general.


Example:

$\displaystyle (\sqrt{-1})(\sqrt{-1}) \ = \ (i)(i) \ = \ i^2 \ = \ -1 \ \ne \ \sqrt{(-1)(-1) \ } \ = \ \sqrt{ \ 1 \ } \ = \ 1$

Last edited by skipjack; November 8th, 2015 at 04:57 PM.
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May 13th, 2015, 12:39 PM   #6
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It's true in the real numbers where this question obviously resides.

MMBt is trying to find errors at the expense of confusing you.

Unless you are working with complex numbers (if you don't know whether you are or not, then you aren't), $\sqrt{a} \times \sqrt{a} = \sqrt{a \times a}$ when $a \ge 0$. In the context of this example, $x$ and $y$ are clearly greater than or equal to zero since $\sqrt x$ and $\sqrt y$ both appear on the left hand side, so they must exist.
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May 16th, 2015, 03:54 PM   #7
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It's true in the real numbers where this question obviously resides.

MMBt is trying to find errors at the expense of confusing you.
v8archie, you are passing off ignorance on this thread as some little child.
I corrected another tutor.

When I make a correction to another tutor about a point, you typed
a lie such as the above.

And think next time, "Gee, the OP didn't make a post referencing what
Math Message Board tutor stated. I won't assume then. I won't fix
something for which there's nothing broken then."

Summation: Get on about your business.

Last edited by Math Message Board tutor; May 16th, 2015 at 04:05 PM.
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May 18th, 2015, 01:47 AM   #8
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Quote:
Originally Posted by Math Message Board tutor View Post
v8archie, you are passing off ignorance on this thread as some little child.
I corrected another tutor.
I'm not a tutor. Just in case you aren't aware, this isn't a tutoring website. It's a maths forum.

Quote:
When I make a correction to another tutor about a point, you typed
a lie such as the above.
I agree with V8Archie on this one.

I agree that $\displaystyle \sqrt{a \times a} \neq \sqrt{a}\sqrt{a}$ if $\displaystyle a$ is negative, as you stated. However, I'm not in the habit of constantly writing absolutely everything down in absolute fine detail just to avoid nitpicking details from pedants.
Thanks from jonah

Last edited by Benit13; May 18th, 2015 at 02:02 AM.
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May 18th, 2015, 05:28 AM   #9
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Let's admit, tacitly, that we are in the real world.

So we have here:

$\displaystyle \sqrt{a}\sqrt{a} = (\sqrt{a})^2 = a
\\\;\\
\sqrt{a^2} = |a|, \forall\ a\in\mathbb{R}$
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May 18th, 2015, 06:30 AM   #10
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Quote:
Originally Posted by Zman15 View Post

I do understand that (x squareroot x - y square root y) ( squareroot x - squaretoot y/(squareroot x)^2 - (squareroot y)^2 is x -y Also will be (x square root x + y square root y)^2 + (square root x - squareoot y)^2/ x -y I also understand that you use square sentence three in this example.
Here is, exactly where it would have its place, a simple algebraic expression:

$\displaystyle E= \dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}-\sqrt{y}}\ \Longrightarrow\ E = \dfrac{(\sqrt{x})^3+(\sqrt{y})^3}{\sqrt{x}-\sqrt{y}}$

Let's write:

$\displaystyle \sqrt{x}=a,\ \ \ \sqrt{y}=b$

Now, we have:

$\displaystyle E=\dfrac{a^3+b^3}{a-b}$

Amplify with a+b.

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