Relate sqrt(4−3x) to sqrt(4.03).

Question

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9. Use the first 3 terms in the expansion of $\sqrt{4-3x}$ to approximate $\sqrt{4.03}$ to 3 decimal places.

Relate $\sqrt{4-3x}$ to $\sqrt{4.03}$ .
We need $4-3x = 4.03$ .
$-3x = 4.03 - 4$ $-3x = 0.03$ $x = \frac{0.03}{-3}$ $x = -0.01$
Write $\sqrt{4-3x}$ in the form $(a+b)^n$ .
$\sqrt{4-3x} = (4-3x)^{1/2}$
Here, $a=4$ , $b=-3x$ , and $n=\frac{1}{2}$ .
Expand using the binomial theorem $(a+b)^n = a^n + n a^{n-1}b + \frac{n(n-1)}{2!} a^{n-2}b^2 + \dots$
The first three terms are:
Term 1: $a^n = 4^{1/2} = 2$
Term 2: $n a^{n-1}b = \frac{1}{2} (4)^{1/2 - 1} (-3x) = \frac{1}{2} (4)^{-1/2} (-3x) = \frac{1}{2} \cdot \frac{1}{\sqrt{4}} (-3x) = \frac{1}{2} \cdot \frac{1}{2

Relate sqrt(4-3x) to sqrt(4.03).