Surd Equation, Please help :)

[onthebeatpete]

Hi everyone, I'm having a bit of trouble with surds. I have gone through a whole heap of questions and thought I was going ok but am having a little trouble understanding a particular question.
Can someone please help me get started or explain how to do this please?

√75 - 2√48 + √98 + 3√243 - 3√12 + 2√72 = a√2 + b√3

I know that If you have

3√2 + 5√2
= 3+5 = 8
= 8√2

Do I need to make my top question to similar numbers as above so i can get to my answer?

Thankyou guys
any help appreciated

 

[Deleted member 4993]

Hi everyone, I'm having a bit of trouble with surds. I have gone through a whole heap of questions and thought I was going ok but am having a little trouble understanding a particular question.
Can someone please help me get started or explain how to do this please?

√75 - 2√48 + √98 + 3√243 - 3√12 + 2√72 = a√2 + b√3

I know that If you have

3√2 + 5√2
= 3+5 = 8....... Do not write this - it is an incorrect statement. In stead do:

3√2 + 5√2

= √2 * (3 + 5)

= 8√2

= 8√2

Do I need to make my top question to similar numbers as above so i can get to my answer?

Thankyou guys
any help appreciated

Yes .... First convert the composite numbers into "square" factors - like -

√75 - 2√48 + √98 + 3√243 - 3√12 + 2√72

= √(5²*3) - 2√(4²*3) + √(7²2) + 3√(9²*3) - 3√(2²*3) + 2√(6²*2)

and continue...

 

[onthebeatpete]

You have been very helpful
Thankyou very much for your help

 

[lookagain]

WHAT are you supposed to do? Sol[v]e for a and b?

Let k = √75 - 2√48 + √98 + 3√243 - 3√12 + 2√72
Then: a√2 + b√3 = k
2 unknowns, so you need 2 equations. So can't be solved. . . . . . Denis, there *are* two equations.

\(\displaystyle 5\sqrt{3} - 2(4)\sqrt{3} + 7\sqrt{2} + 3(9)\sqrt{3} - 3(2)\sqrt{3} + 2(6)\sqrt{2} \ = \ a\sqrt{2} + b\sqrt{3} \ \implies \)
\(\displaystyle 5\sqrt{3} - 8\sqrt{3} + 7\sqrt{2} + 27\sqrt{3} - 6\sqrt{3} + 12\sqrt{2} \ = \ a\sqrt{2} + b\sqrt{3} \ \implies \)
\(\displaystyle 7\sqrt{2} + 12\sqrt{2} \ = \ a\sqrt{2} \ \ \ \) *and* \(\displaystyle \ \ \ 5\sqrt{3} - 8\sqrt{3} + 27\sqrt{3} - 6\sqrt{3} \ = \ b\sqrt{3} \ \ \ \ \ \) Now solve for a and b.

 

[DrPhil]

At 7.13 am, I was non-coffeed and in no shape to "question" anything,
but thought about this during my walk and, and, and....

OK; it's evident that if 19SQRT(2) = aSQRT(2), then a = 19.
BUT:
why is it evident that if 19SQRT(2) + 18SQRT(3) = aSQRT(2) + bSQRT(3),
then a = 19 and b = 18?

(19 - a) sqrt(2) + (18 - b) sqrt(3) = 0
sqrt(3) = (m/n) sqrt(2)

No rational solution for (m/n), except m=n=0

 

[lookagain]

OK; it's evident that if 19SQRT(2) = aSQRT(2), then a = 19.
BUT:
why is it evident that if 19SQRT(2) + 18SQRT(3) = aSQRT(2) + bSQRT(3),
then a = 19 and b = 18? ***

There must have been an assumption that a and b are integers. If a and b don't have to be integers,

then we could have \(\displaystyle \ a = 9\sqrt{6} \ \ and \ \ b = \dfrac{19\sqrt{6}}{3}, \ \ for \ \ example.\)


When you started with the "mumbo-jumbo" equation, you got into a different animal
that isn't really "similar."


Let's take a similar mumbo-jumbo equation that simplifies to:
6SQRT(25) + 3SQRT(16) = aSQRT(25) + bSQRT(16)
Then we jump to conclusions(?) and with audacity announce a = 6 and b = 3 ?

BUT why can't a = 2 and b = 8?
2SQRT(25) + 8SQRT(16) = 10 + 32 = 42
and
6SQRT(25) + 3SQRT(16) = 30 + 12 = 42

Am I missing something?

And 10sqrt(25) + (-2)sqrt(16) = 42

And -6sqrt(25) + 18sqrt(16) = 42

And -14sqrt(25) + 28sqrt(16) = 42

And -18sqrt(25} + 33sqrt(16) = 42

And so on...


(In effect, you have 5a + 6b = 42, which has
an infinite number of integer solutions for
a and b.)

As far as the uniqueness of integer solutions for *** above,
you want proof!?

You can't handle the proof!

(Joke.)

 

[Bob Brown MSEE]

WHAT are you supposed to do? Solbe for a and b?

Let k = √75 - 2√48 + √98 + 3√243 - 3√12 + 2√72
Then: a√2 + b√3 = k
2 unknowns, so you need 2 equations. So can't be solved.

You are correct! Without a limiting assumption, you can NOT get a unique solution.

Lookagain made the assumption that the value of a√2 + b√3 must be a member of an algebraic number field specifically Q(√2,√3). This is reasonable because.
1) No solution for reals (as you point out)
2) (√2,√3) appear in the problem
3) "Surd" appears in the Thread Title.

--------------------

This method is familiar to you in problems involving imaginary number extensions Q(i) to the Rationals (or Reals for that matter).

One equation with two unknowns.
Example: a+bi = (2+i)^2
Result: a=3, b=4

--------------------

Your original objection becomes even more amazing, if we add 1 to k.
Your original objection THEN becomes true! But because it is over constrained.

Let k = √75 - 2√48 + √98 + 3√243 - 3√12 + 2√72 + 1
Then: a√2 + b√3 = k
2 unknowns, so you need 2 equations. But, can't be solved.

Only one equation!!!! But adding one more unknown -- now it has a solution!!!! (as follows)

Let k = √75 - 2√48 + √98 + 3√243 - 3√12 + 2√72 + 1
Then: a√2 + b√3 + c = k
3 unknowns, so you need 3 equations. So CAN be solved.