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# Math : Algebra 101

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Q51  V41
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Re: Math : Algebra 101 [#permalink]
2
Kudos
wow - fantastic work and beautiful formatting!
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Re: Math : Algebra 101 [#permalink]
1
Bookmarks
Thanks, that's cool!
+1
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Re: Math : Algebra 101 [#permalink]
great stufff! kudos
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Re: Math : Algebra 101 [#permalink]
gurpreetsingh wrote:
shrouded1 wrote:
Factorization
This is the easiest approach to solving higher degree equations. Though there is no general rule to do this, generally a knowledge of algebraic identities helps. The basic idea is that if you can write an equation in the form :

$$A*B*C=0$$

where each of A,B,C are algebraic expressions. Once this is done, the solution is obtained by equating each of A,B,C to 0 one by one.

Eg. $$x^3 + 11x + 30 = 0$$ =======> $$x^3 + 11x + 30x = 0$$
$$x * (x^2 + 11x + 30) = 0$$
$$x * (x+5) * (x+6) = 0$$

So the solution is x=0,-5,-6

thanks ! edited ...
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Re: Math : Algebra 101 [#permalink]
Nice!!! Thanks!!!
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Re: Math : Algebra 101 [#permalink]
Good one Shrouded..Thanks !
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Re: Math : Algebra 101 [#permalink]
Nice and thanks!!

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Re: Math : Algebra 101 [#permalink]
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Kudos
Kudos! Thanks for making a valuable contribution .
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Re: Math : Algebra 101 [#permalink]
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Kudos
I do believe you made a mistake in this equation, shouldn't there be 5z instead of z?

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Re: Math : Algebra 101 [#permalink]
More than 2 variables
This is not a case that will be encountered often on the GMAT. But in general for n variables you will need at least n equations to get a unique solution. Sometimes you can assign unique values to a subset of variables using less than n equations using a small trick. For example consider the equations :
x + 2y + 5z = 20
x + 4y + 10z = 40

In this case you can treat 2y+z as a single variable to get :
x + (2y+z) = 20
x + 2*(2y+z) = 40

These can be solved to get x=0 and 2y+z=20

There must be some mistake here because 2y+5z is what equals 20 not 2y+z. If it is correct please tell me how that can happen?

Thanks
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Re: Math : Algebra 101 [#permalink]
mistake corrected ...
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Re: Math : Algebra 101 [#permalink]
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Re: Math : Algebra 101 [#permalink]
is there an easier way to find square root of unusual number ... like 191 etc or to see if it's square root exists or not .. .. any tips and tricks?
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Re: Math : Algebra 101 [#permalink]
simmy818 wrote:
is there an easier way to find square root of unusual number ... like 191 etc or to see if it's square root exists or not .. .. any tips and tricks?

https://www.indianmathonline.com/square.pdf
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Re: Math : Algebra 101 [#permalink]
1
Kudos
Great stuff.
Thank you!
Kudos +1
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Re: Math : Algebra 101 [#permalink]
1
Bookmarks
shrouded1 wrote:
[*]$$(x+y)^2=x^2+y^2+2xy$$
[*]$$(x-y)^2=x^2+y^2-2xy$$
[*]$$x^2-y^2=(x+y)(x-y)$$
[*]$$(x+y)^2-(x-y)^2=4xy$$
[*]$$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$$
[*]$$x^3+y^3=(x+y)(x^2+y^2-xy)$$
[*]$$x^3-y^3=(x-y)(x^2+y^2+xy)$$[/list]

Can somebody give me the link for theory of Inequalities???????
i found the modulus theory in the book but on inequity i couldn't
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Re: Math : Algebra 101 [#permalink]
Need Help!!! Urgent

What is the difference between the solution of below problems, Is there any difference in the approach:-
1. X^3 > X

soln: -1<X<0 or X>1

If we follow the above principle for solving than the answer to below equation must be:

2. a^3< a^2 + 2a

-1<a<0 or a>2

but the answer as per MGMAt quant book is
0<a<2 or a<-1

Pls help me in getting the right concept.........
Re: Math : Algebra 101 [#permalink]
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