May 12th, 2015, 06:25 AM  #1 
Member Joined: Jan 2015 From: Norway Posts: 95 Thanks: 1  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. 
May 12th, 2015, 07:26 AM  #2 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,157 Thanks: 732 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
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 xpart and the ypart: $\displaystyle =x y$ 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: $\displaystyle \frac{x^2 + y^2 + (x+y)\sqrt{y}\sqrt{x}}{xy}$ 
May 12th, 2015, 03:55 PM  #3  
Banned Camp Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191  Quote:
$\displaystyle \sqrt{a}\sqrt{a} \ \ne \ \sqrt{a \times a} $ $\displaystyle \sqrt{a \times a} \ = \ \sqrt{a^2} \ = \ a$ It's because $\displaystyle \ \ \sqrt{a}\sqrt{a} \ = \ (\sqrt{a})^2 \ = \ a.$  
May 12th, 2015, 05:15 PM  #4  
Senior Member Joined: Jan 2014 From: The backwoods of Northern Ontario Posts: 391 Thanks: 70  Quote:
 
May 13th, 2015, 11:26 AM  #5  
Banned Camp Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191  Quote:
Quote:
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.  
May 13th, 2015, 12:39 PM  #6 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2654 Math Focus: Mainly analysis and algebra 
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. 
May 16th, 2015, 03:54 PM  #7  
Banned Camp Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191  Quote:
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.  
May 18th, 2015, 01:47 AM  #8  
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,157 Thanks: 732 Math Focus: Physics, mathematical modelling, numerical and computational solutions  Quote:
Quote:
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. Last edited by Benit13; May 18th, 2015 at 02:02 AM.  
May 18th, 2015, 05:28 AM  #9 
Senior Member Joined: Apr 2014 From: Europa Posts: 584 Thanks: 177  Let's admit, tacitly, that we are in the real world. So we have here: $\displaystyle \sqrt{a}\sqrt{a} = (\sqrt{a})^2 = a \\\;\\ \sqrt{a^2} = a, \forall\ a\in\mathbb{R}$ 
May 18th, 2015, 06:30 AM  #10  
Senior Member Joined: Apr 2014 From: Europa Posts: 584 Thanks: 177  Quote:
$\displaystyle E= \dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}\sqrt{y}}\ \Longrightarrow\ E = \dfrac{(\sqrt{x})^3+(\sqrt{y})^3}{\sqrt{x}\sqrt{y}}$ Let's write: $\displaystyle \sqrt{x}=a,\ \ \ \sqrt{y}=b$ Now, we have: $\displaystyle E=\dfrac{a^3+b^3}{ab}$ Amplify with a+b.  

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