Surd Problem

Simonsky

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Jul 4, 2017
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128
The question is:

The numbers X, Y and Z satisfy Y^2 = XZ

Find Z if X = sq.root 3 and Y = 1-sq.root 3

here's what I did so far: taking the Y value (1-sq.rt 3) and squaring it gave me 1 -2x sq.rt 3 +3 = 4 - 2xsq.rt3 (if I got that right!!)

Here's where I get stuck: so, to get Z, I need a number that multiplies with X (which is sq.rt3) to make Y^2 above.

So: Z = (4-2.sqrt3)/sqrt3 and I got 4/sqrt3 - 2

This is NOT the naswer in my book, so I've gone wrong somewhere maybe in the division process because I divided 2.sqrt3 by sqrt3 and assumed it to be -2

ANy pointers to help me, welcome! Thanks (apologies for doing square root three ways, I eventually settled on sqrt!
 
The question is:

The numbers X, Y and Z satisfy Y^2 = XZ

Find Z if X = sq.root 3 and Y = 1-sq.root 3

here's what I did so far: taking the Y value (1-sq.rt 3) and squaring it gave me 1 -2x sq.rt 3 +3 = 4 - 2xsq.rt3 (if I got that right!!)

Here's where I get stuck: so, to get Z, I need a number that multiplies with X (which is sq.rt3) to make Y^2 above.

So: Z = (4-2.sqrt3)/sqrt3 and I got 4/sqrt3 - 2

This is NOT the naswer in my book, so I've gone wrong somewhere maybe in the division process because I divided 2.sqrt3 by sqrt3 and assumed it to be -2

ANy pointers to help me, welcome! Thanks (apologies for doing square root three ways, I eventually settled on sqrt!

Thought Question: Did you notice that \(\displaystyle 1 -\sqrt{3} < 0\)?

Please read my signature. This may be the problem. \(\displaystyle \dfrac{4-2\sqrt{3}}{\sqrt{3}} = \dfrac{4}{\sqrt{3}}-2 = \dfrac{4\sqrt{3}-6}{3}\) Some teachers will insist on the last version.
 
Thought Question: Did you notice that \(\displaystyle 1 -\sqrt{3} < 0\)?

Please read my signature. This may be the problem. \(\displaystyle \dfrac{4-2\sqrt{3}}{\sqrt{3}} = \dfrac{4}{\sqrt{3}}-2 = \dfrac{4\sqrt{3}-6}{3}\) Some teachers will insist on the last version.


Thanks -not sure where you get the '6' from.

If we've got 4/sqrt3 - 2 and want to get rid of the irrational denominator then we multiply each fraction by sqrt3/sqrt3 ?

So: (3/sqrt3 x sqrt3/sqrt3) - 2x sqrt3/sqrt3 = ( I think) 4sqrt3/3 -2.sqrt3/sqrt3 which leaves us: -2 +4sqrt3/3

Have I missed something?
 
Thanks -not sure where you get the '6' from.
If we've got 4/sqrt3 - 2 and want to get rid of the irrational denominator then we multiply each fraction by sqrt3/sqrt3 ?
So: (3/sqrt3 x sqrt3/sqrt3) - 2x sqrt3/sqrt3 = ( I think) 4sqrt3/3 -2.sqrt3/sqrt3 which leaves us: -2 +4sqrt3/3
Have I missed something?
\(\displaystyle \begin{align*}\dfrac{4\sqrt{3}}{\sqrt{3}}-2&=\dfrac{4-2\sqrt{3}}{\sqrt{3}}\\&=\dfrac{4-2\sqrt{3}}{\sqrt{3}}\left(\dfrac{\sqrt3}{\sqrt{3}}\right)\\&= \dfrac{4\sqrt{3}-6}{3} \end{align*}\)
 
\(\displaystyle \begin{align*}\dfrac{4\sqrt{3}}{\sqrt{3}}-2&=\dfrac{4-2\sqrt{3}}{\sqrt{3}}\\&=\dfrac{4-2\sqrt{3}}{\sqrt{3}}\left(\dfrac{\sqrt3}{\sqrt{3}}\right)\\&= \dfrac{4\sqrt{3}-6}{3} \end{align*}\)

But haven't you taken the whole thing as divisible by sqrt3 ? isn't the two a separate number that you need to multiply by sqrt3/sqrt3 separately?
 
XZ = Y^2 (GIVEN)

Z = Y^2 / X

Z = (1 - √3)^2 / √3

Z = (1 - 2√3 + 3) / √3

Finish it...
Denis, the OP got one step further than that by realizing 1+ 3 = 4 and Z = (4 - 2√3) / √3
 
But haven't you taken the whole thing as divisible by sqrt3 ? isn't the two a separate number that you need to multiply by sqrt3/sqrt3 separately?
The above reply begs the question: do you understand the operations with fractions?
\(\displaystyle \dfrac{A}{B}-C=\dfrac{A-BC}{B}\)
\(\displaystyle \dfrac{A}{B}-C=\dfrac{A-BC}{B}\left(\dfrac{B}{B}\right)=\dfrac{AB-B^2C}{B^2}\)
 
But haven't you taken the whole thing as divisible by sqrt3 ? isn't the two a separate number that you need to multiply by sqrt3/sqrt3 separately?
No, that is not correct. You are saying that (3*4)*5 = (3*5)*(4*5) which is not true. If you choose not to multiply the 3 and 4 first then you should get (3*5)*4 OR (4*5)*3
 
\(\displaystyle \begin{align*}\dfrac{4\sqrt{3}}{\sqrt{3}}-2&=\dfrac{4-2\sqrt{3}}{\sqrt{3}}\\&=\dfrac{4-2\sqrt{3}}{\sqrt{3}}\left(\dfrac{\sqrt3}{\sqrt{3}}\right)\\&= \dfrac{4\sqrt{3}-6}{3} \end{align*}\)
But haven't you taken the whole thing as divisible by sqrt3 ? isn't the two a separate number that you need to multiply by sqrt3/sqrt3 separately?
Are you serious?

\(\displaystyle (4 - 2a) * a = 4a - (2 * a * a) = 4a - 2a^2,\) right?

\(\displaystyle a = \sqrt{3} \implies (4 - 2\sqrt{a}) * \sqrt{a} = 4\sqrt{3} - 2(\sqrt{3})^2 = 4\sqrt{3} - 2 * 3 = 4\sqrt{3} - 6.\)
 
Guess that's why he wasn't too impressed with my post:(

Not at all!

I'm just struggling with the fractions. I still don't get how BOTH numbers become part of one fraction-I'm afraid I don't understand the examples given to illustrate this. Can anyone help here? I don't really get the A/B - C = (A-BC)/B

Thanks for help.
 
This is NOT the naswer in my book, so I've gone wrong somewhere maybe in the division process because I divided 2.sqrt3 by sqrt3 and assumed it to be -2

ANy pointers to help me, welcome! Thanks (apologies for doing square root three ways, I eventually settled on sqrt!
In the future you can find the difference between your answer and the books answer and if the difference is 0 then both answers are the same. This way, if you get 0 as the difference, you know that your answer is correct and you can spend your time figuring out how to get your answer to look like the one in the book. This will always result in multiplying by 1 in a unique fashion (like sqrt(3)/ sqrt(3) or7/7 or 19/19 ...)
 
Not at all!

I'm just struggling with the fractions. I still don't get how BOTH numbers become part of one fraction-I'm afraid I don't understand the examples given to illustrate this. Can anyone help here? I don't really get the A/B - C = (A-BC)/B

Thanks for help.
Some useful basic rules.

\(\displaystyle a * 1 \equiv a \equiv 1 * a.\) Rule 1.

\(\displaystyle \dfrac{b}{1} \equiv b.\) Rule 2.

\(\displaystyle c \ne 0 \implies \dfrac{c}{c} = 1.\) Rule 3.

\(\displaystyle r \ne 0 \implies \dfrac{s}{r} \pm \dfrac{t}{r} = \dfrac{s \pm t}{r}.\) Rule 4.

\(\displaystyle j \ne 0 \ne k \implies \dfrac{m}{j} * \dfrac{n}{k} = \dfrac{mn}{jk}.\) Rule 5.

Any problem with any of those rules?

Now let's see.

\(\displaystyle w = \dfrac{x}{y} - z.\)

\(\displaystyle \therefore w = \dfrac{x}{y} - \dfrac{z}{1} \text { by Rule 2.}\)

\(\displaystyle \therefore w = \dfrac{x}{y} - \left ( \dfrac{z}{1} * 1 \right ) \text { by Rule 1.}\)

\(\displaystyle \therefore w = \dfrac{x}{y} - \left ( \dfrac{z}{1} * \dfrac{y}{y} \right ) \text { by Rule 3.}\)

\(\displaystyle \therefore w = \dfrac{x}{y} - \dfrac{z * y}{1 * y} \text { by Rule 5.}\)

\(\displaystyle \therefore w = \dfrac{x}{y} - \dfrac{yz}{y} \text { by Rule 1.}\)

\(\displaystyle \therefore w = \dfrac{x - yz}{y} \text { by Rule 4.}\)
 
Some useful basic rules.

Thanks Jeff. I get it! I now realise it is just the normal process of getting the same denominator when you have addition/subtraction of fractions which i thought I knew. It just LOOKED different in those examples!

I find a slight difference in visuals can throw me!

Thanks.
 
Some useful basic rules.

Thanks Jeff. I get it! I now realise it is just the normal process of getting the same denominator when you have addition/subtraction of fractions which i thought I knew. It just LOOKED different in those examples!

I find a slight difference in visuals can throw me!

Thanks.
No problem. The problem with TEACHING math is that steps that have become obvious through experience to the teacher are left out.

I see \(\displaystyle w = \dfrac{x}{y} - z\), and I "see" that \(\displaystyle \dfrac{x}{y} - z \equiv \dfrac{x - yz}{y}.\)

A BIG part of teaching math is that you must NOT forget that what is obvious to you may not be obvious to whomever you are trying to teach. I screw up there a lot. It is why when I tutor in person I always look at the tutee's face. I am looking for that look of utter confusion that signals I have once again failed as a tutor.
 
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