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Year 12s: what will be the first way you research your future options? >>

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Taking out negative factors

for example

2x(5x-1)^{\frac{1}{2}} +\dfrac{5x^2}{2} (5x-1)^{-\frac{1}{2}}

how do i take out a factor of that (5x-1) what power of it do i take out?

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Original post by will'o'wisp

for example

2x(5x-1)^{\frac{1}{2}} +\dfrac{5x^2}{2} (5x-1)^{-\frac{1}{2}}

how do i take out a factor of that (5x-1) what power of it do i take out?

Depends what form you want it in

I'd take out the one to negative half

OP

Original post by RDKGames

Depends what form you want it in

I'd take out the one to negative half

oh so how would i take out the negative half then?

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Original post by will'o'wisp

oh so how would i take out the negative half then?

The second term already has it as part of the product so you can easily deal with that. Since the first term doesn't have it, you'd need to divide that term by the factor from which point you can simplify the bracket.

(edited 7 years ago)

OP

Original post by RDKGames

The second term already has it as part of the product so you can easily deal with that. Since the first term doesn't have it, you'd need to multiply that term by the factor from which point you can simplify the bracket.

so for the left part of the (5x-1), if i take out the -0.5 i'll get (5x-1)^-1 ?

Study Forum Helper

Original post by will'o'wisp

so for the left part of the (5x-1), if i take out the -0.5 i'll get (5x-1)^-1 ?

Sorry can you please word it using some correct terminology? I find it hard to understand what you're asking here... :s

OP

Original post by RDKGames

Sorry can you please word it using some correct terminology? I find it hard to understand what you're asking here... :s

referring specifically to the stuff of the left side of the equation

2x(5x-1)^{\frac{1}{2}}

(5x-1)^{-\frac{1}{2}}

is taken out then with the stuff on the left am i left with

2x(5x-1)^{-1}

Study Forum Helper

Original post by will'o'wisp

referring specifically to the stuff of the left side of the equation

2x(5x-1)^{\frac{1}{2}}

(5x-1)^{-\frac{1}{2}}

is taken out then with the stuff on the left am i left with

2x(5x-1)^{-1}

No you would be left with +1 as the exponent. You can easily check this yourself by expanding the brackets again and getting what you started with.

(edited 7 years ago)

OP

Original post by RDKGames

No you would be left with +1 as the exponent. You can easily check this yourself by expanding the brackets again and getting what you started with.

ok thanks

Original post by will'o'wisp

oh so how would i take out the negative half then?

To take out a factor, you need to create that factor in each term if it is not currently present. For example, consider factorising

a^{-1/2}

as follows:

a+a^{-\frac{1}{2}} = b a^{-\frac{1}{2}} + 1 \times a^{-\frac{1}{2}} = a^{-\frac{1}{2}}(b+1)

where we want

b a^{-\frac{1}{2}} = a \Rightarrow b = \frac{a}{a^{-\frac{1}{2}}} = a^{\frac{3}{2}}

. So:

a+a^{-\frac{1}{2}} = a^{-\frac{1}{2}}(a^\frac{3}{2}+1)

Or similarly and a bit quicker:

a^2+a+1 = a^2+a^1+a^0 = \\ \\ a^{2+\frac{1}{2}}a^{-\frac{1}{2}} + a^{1+\frac{1}{2}}a^{-\frac{1}{2}} + a^{0+\frac{1}{2}}a^{-\frac{1}{2}} = a^{-\frac{1}{2}}(a^\frac{5}{2} +a^\frac{3}{2}+a^\frac{1}{2})

OP

Original post by atsruser

To take out a factor, you need to create that factor in each term if it is not currently present. For example, consider factorising

a^{-1/2}

as follows:

a+a^{-\frac{1}{2}} = b a^{-\frac{1}{2}} + 1 \times a^{-\frac{1}{2}} = a^{-\frac{1}{2}}(b+1)

where we want

b a^{-\frac{1}{2}} = a \Rightarrow b = \frac{a}{a^{-\frac{1}{2}}} = a^{\frac{3}{2}}

. <---- I don't get this line
So:

a+a^{-\frac{1}{2}} = a^{-\frac{1}{2}}(a^\frac{3}{2}+1)

Or similarly and a bit quicker:

a^2+a+1 = a^2+a^1+a^0 = \\ \\ a^{2+\frac{1}{2}}a^{-\frac{1}{2}} + a^{1+\frac{1}{2}}a^{-\frac{1}{2}} + a^{0+\frac{1}{2}}a^{-\frac{1}{2}} = a^{-\frac{1}{2}}(a^\frac{5}{2} +a^\frac{3}{2}+a^\frac{1}{2})

^^ as shown above

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