Question

If \[x\] is small, \[\sqrt {{x^2} + 16} \, - \,\sqrt {{x^2} + 9} \] is equal to
A) \[1 + \dfrac{{{x^2}}}{{24}}\]
B) \[1 - \dfrac{{{x^2}}}{{24}}\]
C) \[1 + \dfrac{{{x^2}}}{{48}}\]
D) \[1 - \dfrac{{{x^2}}}{{48}}\]

Answer 1

Hint: We can solve this sum by considering the square root as power of \[\dfrac{1}{2}\] for each term and expand using Binomial expansion for fractions. We will have to neglect the higher powers when using Binomial expansion. This is because it is given in the question that \[x\] is small. After doing these we obtain the required answer.

Formula used: Binomial expansion for fractional powers
\[{(1 + x)^\alpha }\, = \,1 + \alpha x\, + \dfrac{{(\alpha )(\alpha - 1)}}{{2!}}{x^2} + .....\] Where \[\alpha = \,\dfrac{p}{q}\]

Complete step-by-step answer:
It is given that the equation, \[\sqrt {{x^2} + 16} \, - \,\sqrt {{x^2} + 9} \],
We start by considering the square root of each term as a power of \[\dfrac{1}{2}\] for that term.
That is, the given equation can be written as \[{({x^2} + 16)^{\dfrac{1}{2}}}\, - \,{({x^2} + 9)^{\dfrac{1}{2}}}\]
Now we can reduce the above as\[{(16)^{\dfrac{1}{2}}}{\left( {1 + \dfrac{{{x^2}}}{{16}}} \right)^{\dfrac{1}{2}}}\, - \,{(9)^{\dfrac{1}{2}}}\left( {1 + \dfrac{{{x^2}}}{9}} \right) \to (A)\].
We have taken 16 as a common factor from the first term which will result in 16 dividing the \[{x^2}\] and similarly for the second term, we take 9 as a common factor.
So we can write it as, \[{(16)^{\dfrac{1}{2}}} = \sqrt {16} = 4\]and we have\[{(9)^{\dfrac{1}{2}}} = \sqrt 9 = 3\],
We can further reduce equation (A) using these as,
\[ \Rightarrow 4{\left( {1 + \dfrac{{{x^2}}}{{16}}} \right)^{\dfrac{1}{2}}}\, - \,3{\left( {1 + \dfrac{{{x^2}}}{9}} \right)^{\dfrac{1}{2}}} \to (1)\].
We consider this reduced form as equation (1).
Now we use binomial expansion for fractional powers, we know that
\[{(1 + x)^\alpha }\, = \,1 + \alpha x\, + \dfrac{{(\alpha )(\alpha - 1)}}{{2!}}{x^2} + .....\] Where \[\alpha = \,\dfrac{p}{q}\]
In the Problem that we have to solve, \[\alpha = \dfrac{1}{2}\]
Now, we can use the formula and expand the equation (1) one by one,
\[ \Rightarrow {\left( {1 + \dfrac{{{x^2}}}{{16}}} \right)^{\dfrac{1}{2}}}\, = 1 + \dfrac{1}{2} \times \dfrac{{{x^2}}}{{16}} + \dfrac{{\dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{{{x^4}}}{{{{16}^2}}}}}{{2!}} + ...........\]
Also, we expand \[{\left( {1 + \dfrac{{{x^2}}}{9}} \right)^{\dfrac{1}{2}}} = 1 + \dfrac{1}{2} \times \dfrac{{{x^2}}}{9} + \dfrac{{\dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{{{x^4}}}{{{9^2}}}}}{{2!}} + ...........\]
Since it is given that \[x\]is small, we can neglect the higher powers of x.
By using the above two expansions in equation (1), we get,
\[ \Rightarrow 4\left( {1 + \dfrac{{{x^2}}}{{32}}} \right)\, - 3\,\left( {1 + \dfrac{{{x^2}}}{{18}}} \right)\]
On multiplying the term and we get,
\[ \Rightarrow 4 + \dfrac{{4{x^2}}}{{32}} - 3 - \dfrac{{3{x^2}}}{{18}}\]
We multiple 4 and 3 inside the respective terms and then 32 is a multiple of 4 and 18 is a multiple of 3, We arrive at,
\[ \Rightarrow 4 - 3 + \dfrac{{{x^2}}}{8} - \dfrac{{{x^2}}}{6}\]
On subtracting we get,
\[ \Rightarrow 1 + \dfrac{{{x^2}}}{8} - \dfrac{{{x^2}}}{6}\]
To simplify the above expression, we can take LCM of 8 and 6, Which will be 24 and it will further reduce as follows
\[ \Rightarrow 1 + \dfrac{{3{x^2} - \,4{x^2}}}{{24}}\]
Let us subtract the numerator term and we get,
\[ \Rightarrow 1 - \dfrac{{\,{x^2}}}{{24}}\]
Therefore, If \[x\] is small, \[\sqrt {{x^2} + 16} \, - \,\sqrt {{x^2} + 9} \] is equal to \[1 - \dfrac{{\,{x^2}}}{{24}}\].

Option B is the correct answer.

Note: In this type of question it is important to understand how we use the Binomial expansion. That is, we take the value of x in the Binomial expansion as \[\dfrac{{{x^2}}}{{16}}\] for the first term and \[\dfrac{{{x^2}}}{9}\] for the second term.
A fraction exponent, that is \[\dfrac{1}{n}\] which means that we can take the ${n^{th}}$ root,
Now we can write it as, $\sqrt[n]{x} = {x^{\dfrac{1}{n}}}$