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    i have been stuck on this question for 2 days please someone out their help me out of my misery. part c

    3b. binomial expansion of (1+1.5)^0.5= 1+0.75x-9/32x^2

    3c. hence show that ((2+3x)/8)^0.5 approx.= a +bx+cx^2
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    Would love to help but hard to read question written like that anyway you can go online and copy the question?
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    (Original post by whizz-kid)
    i have been stuck on this question for 2 days please someone out their help me out of my misery. part c

    3b. binomial expansion of (1+1.5)^0.5= 1+0.75x-9/32x^2

    3c. hence show that ((2+3x)/8)^0.5 approx.= a +bx+cx^2
    It needs factorising and then you will get the second part, like this; (x/8)^0.5=(x/2)^0.5(1/4)^0.5=(1/2)(x/2)^0.5.
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    (Original post by jameswhughes)
    It needs factorising and then you will get the second part, like this; (x/8)^0.5=(x/2)^0.5(1/4)^0.5=(1/2)(x/2)^0.5.
    i dont understand in the markscheme it says: 0.5+(3/8)x + (9/64)^2
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    (Original post by whizz-kid)
    i dont understand in the markscheme it says: 0.5+(3/8)x + (9/64)^0.5
    Yeah, that's right. What I was trying to say is when you take the quarter out it comes out as a half, because of the square root, so the new answer is half of what you had for the first answer.
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    (Original post by jameswhughes)
    Yeah, that's right. What I was trying to say is when you take the quarter out it comes out as a half, because of the square root, so the new answer is half of what you had for the first answer.
    thank you for help i understand it finally getting some head way
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    (Original post by whizz-kid)
    thank you for help i understand it finally getting some head way
    ok, good
 
 
 
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Updated: April 12, 2011
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