Question

I need help using Pascal's triangle to expand the expression: (1/x - √x)^5

Answer 1

this is the equation: \[(\frac{ 1 }{ x } - \sqrt{x})^{5}\]

Answer 2

http://www.mathsisfun.com/images/pascals-triangle-2.gif

Answer 3

@jdoe0001 i got that. i'm confused on how to calculate the square roots after expanding the equation

Answer 4

do you know about combinations..?

Answer 5

@campbell_st i know \[(a+b)^{5} = a ^{5}+5a ^{4}b+10a ^{3}b ^{2}+10a ^{2}b ^{3}+5ab ^{4}+b ^{5}\]

Answer 6

because combinations are used to get to coefficients of each term

\[(a + b)^5 = ^5C_{0} (a)^{5 - 0}(b)^0 + ^5C_{1}a^{5 -1}{b}^1 +.....\]

in general its

\[(a + b)^n = \sum {^nC_{r} (a)^{n - r}b^r}\]

where r = 0, 1, 2, 3, 4, ... n

Answer 7

that way you don't need to worry about pascals triangle

and you should write your terms in index form... easier to calculate

\[(x^{-1} - x^{\frac{1}{2}})^5 = ^5C_{0}(x^{-1})^5(x^{\frac{1}{2}})^0 + ^5C_{1} (x^{-1})^{5 -1}(x^{\frac{1}{2}})^1+...\]

Answer 8

so, what part is the one you need help with?

Answer 9

it's not the coefficients, I can see that

Answer 10

@jdoe0001 i got \[\frac{ 1 }{ x ^{5} }+\frac{ 5 }{ x ^{4} }(-\sqrt{x})+\frac{ 10 }{x ^{3} }(-\sqrt{x})^{2}+\frac{ 10 }{ x ^{2} }(-\sqrt{x})^{3}+\frac{ 5 }{ x }(-\sqrt{x})^{4}+(-\sqrt{x})^{5}\]

Answer 11

@jdoe0001 i got lost right there

Answer 12

ok

Answer 13

keep it in index form until you have a final term

\[x^-4 \times x^{\frac{1}{2}} = x^{-\frac{7}{2}} = \frac{1}{\sqrt{x^7}}\]

Answer 14

the answer is \[\frac{ 1 }{ x ^{5} }-\frac{ 5 }{ x ^{\frac{ 7 }{ 2 }} }+\frac{ 10 }{ x ^{2} }+\frac{ 10 }{ x ^{\frac{ 1 }{ 2 }} }+5x-x ^{\frac{ 5 }{ 2 }}\] but i don't know how to get it from there

Answer 15

nope... because you would need to use your index laws skills to write the terms correctly...

Answer 16

$$
\cfrac{1}{x^5}+\cfrac{5\times -\sqrt{x}}{x^5} \implies \cfrac{1+(x(-5\sqrt{x}))}{x^5}\\
\boxed{\cfrac{1-5x\sqrt{x}}{x^5}}
$$

Answer 17

and I think you have the sign of the 4th term incorrect...

Answer 18

which is just plain fraction addition :|

Answer 19

i was just writing what my book wrote. i know the fourth term should be negative/minus

Answer 20

$$
\cfrac{1}{x^5}+\cfrac{5\times -\sqrt{x}}{x^4} \implies \cfrac{1+(x(-5\sqrt{x}))}{x^5}\\
\boxed{\cfrac{1-5x\sqrt{x}}{x^5}}
$$ had a typo on an exponent :(

Answer 21

well if the book allows factional indices... then thats what you write

Answer 22

@campbell_st i never learned the index form in class. how do you do it?

Answer 23

wow... its really hard to do binomial theorem work without index law skills... its a fundamental

here is a link with some explanations. Its basically a Year 8 piece of work.

http://www.staff.vu.edu.au/mcaonline/units/indices/indexlaws.html

http://mathematics.laerd.com/maths/indices-intro.php

hope this helps

Answer 24

@campbell_st thank you

Answer 25

basically, you'd end up just working primarily with the roots and values of x, for say as example in the 5th term:
\(
(- \sqrt{x^4}) \implies (-1 \times \sqrt{x^4}) \implies -1 \times x^{4/2} \implies -x^2
\)

Answer 26

@jdoe0001 i barely got that haha, thank you so much