Question on binomial theorem

Question:
Find the first four terms of the expansion of $(1-8x)^{0.5}$ in ascending of x. Substitute $x = \frac{1}{100}$ and obtain the value of $\sqrt{23}$ correct to five significant figures

a) find the first four terms of the expansion of $(1-8x)^{0.5}$ in ascending of x.
$(1-8x)^{0.5} = \left ( 1 + ((0.5)(-8x))+\frac{(0.5)(-0.5)(-8x)^{2}}{1 \times 2} + \frac{(0.5)(-0.5)(-1.5)(-8x)^{3}}{1 \times 2 \times 3} \right )$
$\left ( 1-4x -8x^2 - 32x^3 \right )$

b)Substitute and obtain the value of $\sqrt{23}$ correct to five significant figures
I have problem with this part

now it says to sub $x = \frac{1}{100}$
so; $\left ( 1 - (8\frac{1}{100}) \right )^{0.5} = 0.9591663047$

using the expansion;

$\left ( 1-4x -8x^2 - 32x^3 \right )$

i sub x = 1/00 into the expansion

$\left ( 1-\left (4 \left(\frac{1}{100}\right) \right) -\left (8\left(\frac{1}{100}\right)^{2} \right) - \left ( 32\left ( \frac{1}{100} \right )^{3} \right ) \right )$

it produces 0.959168
i know that $\sqrt{23} =4.795831523$

My issue is that I dont know why I am not getting 4.795831523
since part b is a simple substitution question asking me to sub x = 1/100 shouldn't the I get the answer 4.795831523, when i sub x= 1/100 into the expansion or binomial.

thank you for helping me

Question:
Find the first four terms of the expansion of $(1-8x)^{0.5}$ in ascending of x. Substitute $x = \frac{1}{100}$ and obtain the value of $\sqrt{23}$ correct to five significant figures

a) find the first four terms of the expansion of $(1-8x)^{0.5}$ in ascending of x.
$(1-8x)^{0.5} = \left ( 1 + ((0.5)(-8x))+\frac{(0.5)(-0.5)(-8x)^{2}}{1 \times 2} + \frac{(0.5)(-0.5)(-1.5)(-8x)^{3}}{1 \times 2 \times 3} \right )$
$\left ( 1-4x -8x^2 - 32x^3 \right )$

b)Substitute and obtain the value of $\sqrt{23}$ correct to five significant figures
I have problem with this part

now it says to sub $x = \frac{1}{100}$
so; $\left ( 1 - (8\frac{1}{100}) \right )^{0.5} = 0.9591663047$

using the expansion;

$\left ( 1-4x -8x^2 - 32x^3 \right )$

i sub x = 1/00 into the expansion

$\left ( 1-\left (4 \left(\frac{1}{100}\right) \right) -\left (8\left(\frac{1}{100}\right)^{2} \right) - \left ( 32\left ( \frac{1}{100} \right )^{3} \right ) \right )$

it produces 0.959168
i know that $\sqrt{23} =4.795831523$

My issue is that I dont know why I am not getting 4.795831523
since part b is a simple substitution question asking me to sub x = 1/100 shouldn't the I get the answer 4.795831523, when i sub x= 1/100 into the expansion or binomial.

thank you for helping me

X = 1/100 does not produce sqrt(23) for
(1 - 8x)^0.5
Just sub in the value?

Can you transform
(1 - 8/100)^0.5
Into sqrt(23)?

yes i can i get $\left ( \frac{23}{25} \right )^{0.5}$
i see now i thought that binomial theorem can be used to find approx but this question is not asking me the approximate to the square roof of 23
i thought that it was an approximate type of question. I personally do not see the point of part b and what it is trying to teach me

yes i can i get $\left ( \frac{23}{25} \right )^{0.5}$
i see now i thought that binomial theorem can be used to find approx but this question is not asking me the approximate to the square roof of 23
i thought that it was an approximate type of question. I personally do not see the point of part b and what it is trying to teach me

You can only have worked out the value of the LHS when x = 1/100 with a calculator. This defeats the object of the question - the idea of binomial approximations in this context is to find the approximate value of roots by using only addition and multiplication of rational numbers.

When you substitute x = 1/100 into the LHS, you get (1 - 8/100)^0.5 = (1 - 2/25)^0.5 = (23/25)^0.5. Does that give you any ideas?

$(23/25)^0.5$ can be changed to $\left ( 1-\left (4 \left(\frac{1}{100}\right) \right) -\left (8\left(\frac{1}{100}\right)^{2} \right) - \left ( 32\left ( \frac{1}{100} \right )^{3} \right ) \right )$

then to transfer to $\sqrt{23}$ by multiplying $\left ( 1-\left (4 \left(\frac{1}{100}\right) \right) -\left (8\left(\frac{1}{100}\right)^{2} \right) - \left ( 32\left ( \frac{1}{100} \right )^{3} \right ) \right )$ by 5

thus;

$5 \times \left ( 1-\left (4 \left(\frac{1}{100}\right) \right) -\left (8\left(\frac{1}{100}\right)^{2} \right) - \left ( 32\left ( \frac{1}{100} \right )^{3} \right ) \right )$

yes i can i get $\left ( \frac{23}{25} \right )^{0.5}$
i see now i thought that binomial theorem can be used to find approx but this question is not asking me the approximate to the square roof of 23
i thought that it was an approximate type of question. I personally do not see the point of part b and what it is trying to teach me