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Hi, I am having trouble figuring out how to do these types of problems,
(p + 5)^(2/3) + 2(p + 5)^(1/3) - 15 = 0
Can someone explain this to me?
2 Replies - 454 Views - Last Post: 02 December 2010 - 12:46 AM
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lordofduct
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Re: Math Problem
Posted 02 December 2010 - 12:46 AM
fractional exponents are roots
and with out the /3 that's a standard binomial: ax^2 + bx + c
the factor of a binomial is (x + A)(x + B )
lastly we know the distributive property of exponents: a^x * b^x = (ab)^x
thing about in the factoring... we have a variable there for x. variables can represent ANY value... so imagine if x == u^k... and lets say that u = (p+5) and k = 1/3... what happens then?
your equation up there becomes:
(p + 5)^(2/3) + 2(p + 5)^(1/3) - 15 = 0
replace with u and k
u^2k + 2u^k - 15 = 0
replace u and k with x
x^2 + 2x - 15 = 0
factor... then replace the x back with what it aught to be.
(x + 5)(x - 3) = 0
(u^k + 5)(u^k - 3) = 0
((p+5)^1/3 + 5)((p+5)^1/3 - 3) = 0
I know I gave the answer, but I was showing you the work and process of how you should look at it. This is algebra, it's about recognizing the ambiguity of variables. They are to represent ALL REAL NUMBERS, all the algebraic theorems are based under this idea... so thusly you can create these "place holders" as such. That's the whole point of algebra...
[edit]
I forgot to point out...
I said that a variable represents real numbers, and there in lies a catch. If you're defining your own variable as a complex value underneath, the value must be functional. This actually is the point of all that 'functions' stuff in algebra... in this this example I defined x actually as:
x = f(p) = (p + 5)^1/3
f(p) MUST be functional for it to work (otherwise the result isn't always a real number and in turn becomes undefined). The fact this question uses cube root is very special, cuberoots are functional... where as squareroots (or even roots in general) are not functional. Others to be wary of are division by a variable (eg: x = f(p) = 3/(p-1), because when p = 1, we get divided by 0). Keep in mind, to be functional for ANY and ALL inputs there is ONE output. Not two, not infinite, and certainly not undefined.
Of course down the line in later classes, where substitution like this becomes a bigger necessity you'll learn about factoring and reducing equations in specific domains. Basically stating that some equation factors to these values as long as your between some given values.
and with out the /3 that's a standard binomial: ax^2 + bx + c
the factor of a binomial is (x + A)(x + B )
lastly we know the distributive property of exponents: a^x * b^x = (ab)^x
thing about in the factoring... we have a variable there for x. variables can represent ANY value... so imagine if x == u^k... and lets say that u = (p+5) and k = 1/3... what happens then?
your equation up there becomes:
(p + 5)^(2/3) + 2(p + 5)^(1/3) - 15 = 0
replace with u and k
u^2k + 2u^k - 15 = 0
replace u and k with x
x^2 + 2x - 15 = 0
factor... then replace the x back with what it aught to be.
(x + 5)(x - 3) = 0
(u^k + 5)(u^k - 3) = 0
((p+5)^1/3 + 5)((p+5)^1/3 - 3) = 0
I know I gave the answer, but I was showing you the work and process of how you should look at it. This is algebra, it's about recognizing the ambiguity of variables. They are to represent ALL REAL NUMBERS, all the algebraic theorems are based under this idea... so thusly you can create these "place holders" as such. That's the whole point of algebra...
[edit]
I forgot to point out...
I said that a variable represents real numbers, and there in lies a catch. If you're defining your own variable as a complex value underneath, the value must be functional. This actually is the point of all that 'functions' stuff in algebra... in this this example I defined x actually as:
x = f(p) = (p + 5)^1/3
f(p) MUST be functional for it to work (otherwise the result isn't always a real number and in turn becomes undefined). The fact this question uses cube root is very special, cuberoots are functional... where as squareroots (or even roots in general) are not functional. Others to be wary of are division by a variable (eg: x = f(p) = 3/(p-1), because when p = 1, we get divided by 0). Keep in mind, to be functional for ANY and ALL inputs there is ONE output. Not two, not infinite, and certainly not undefined.
Of course down the line in later classes, where substitution like this becomes a bigger necessity you'll learn about factoring and reducing equations in specific domains. Basically stating that some equation factors to these values as long as your between some given values.
This post has been edited by lordofduct: 02 December 2010 - 01:37 AM
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