# Last C4 Series Q.

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#1
Okay, pretty much covered this topic, just the last one I am stuck on;

20.)
a.)
Expand (1+2x)^(-1/2), |x|<(1/2), in ascending powers of x up to and including the term in xΒ³.

b.)
Hence, show that for small values of x,

2-5x/√(1+2x) ≈ 2-7x+8xΒ²-(25/2)xΒ³

c.)
Solve the equation;

2-5x/√(1+2x) = √3.

d.)
Use answers to b & c to find an approximate value for √3.

0
15 years ago
#2
2-5x/√(1+2x) = √3
or
(2-5x)/√(1+2x) = √3
?
0
#3
(Original post by mik1w)
2-5x/√(1+2x) = √3
or
(2-5x)/√(1+2x) = √3
?
(2-5x)/√(1+2x) = √3
0
#4
Any takers?
0
15 years ago
#5
(Original post by Slice'N'Dice)
(2-5x)/√(1+2x) = √3
(2-5x)/√(1+2x) = √3

put y = 1+2x

(2-5x)/√y = √3
2/√y - 5x/√y = √3
2 - 5x = √3√y
2 - 5x = √(3y)
2 - 5x = √(3 + 6x)
25x^2 - 20x + 4 = 3 + 6x
25x^2 - 26x + 1 = 0
(25x - 1)(x - 1) = 0
x = 1 or 1/25

put both solutions back in to check that neither are "false" from the squaring we did earlier, that should be okay.
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#6
Alrighty will do. Thankyou m.
0
15 years ago
#7
I haven't learnt the rest. only upto C3 so far
0
15 years ago
#8
(Original post by Slice'N'Dice)
Okay, pretty much covered this topic, just the last one I am stuck on;

20.)
a.)
Expand (1+2x)^(-1/2), |x|<(1/2), in ascending powers of x up to and including the term in xΒ³.

b.)
Hence, show that for small values of x,

2-5x/√(1+2x) ≈ 2-7x+8xΒ²-(25/2)xΒ³

c.)
Solve the equation;

2-5x/√(1+2x) = √3.

d.)
Use answers to b & c to find an approximate value for √3.

(1+2x)^(-1/2)=1-x+(-1/2)(-3/2)/2!(2x)^2+(-1/2)(-3/2)(-5/2)/3!(2x^3)
=1-x+3x^2/2-5/2x^3
for small values
2-5x/√(1+2x)=(2-5x)(1-x+3x^2/2-5/2x^3)
=2-2x+3x^2-5x^3-5x+5x^2-15/2x^3
=2-7x+8x^2-(25/2)x^3..............(1)
ill take the 1/25 as a correct solution for 2-5x/√(1+2x) = √3.
so approx for √3 given by
2-(7/25)+8/625-1/1250 .............using (1) with x=1/25
=(2500-350+16-1)/1250
=2165/1250
=433/250
=1.732
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