Question

Find the square root of the following (correct to three decimal places).
(i) 7
(ii) 2.5

Answer 1

Hint: To solve this question, we will use the concept of differentiation. First we will find the perfect square near the number, whose we have to evaluate the square root. Then, we will find the change between the number we get as perfect square and number whose square value we have to evaluate. Hence, then using some differential formula we will evaluate the value of the square root of 7 and 2.5.

Complete step by step answer:
Let, y be the function of $\sqrt{x}$, and let $x+\delta x$ denotes the value of number of which we have to evaluate the value of square root, where $\delta x$ denotes change in y and number $x+\delta x$. Then using the differential, we will evaluate the value of dy which denotes change in y when there is change of dx in x.
So, we get the relation $\sqrt{x+\delta x}=y+dy$.
(i) Now, firstly we will solve for finding the square root of 7.
Now, nearest number to 7 whose square root can be taken is 9, as $\sqrt{9}=3$ , so let us consider that x = 9 and $\delta x=dx=-2$ , as $x+\delta x=9+(-2)=7$
Now, the square root of x will lead to any number equals to say, y.
So, consider $y=\sqrt{x}$……( i );
As y is dependent on x so x is independent variable and y is dependent variable.
Differentiating ( i ) with respect to x, we get
$\dfrac{dy}{dx}=\dfrac{d}{dx}\left( \sqrt{x} \right)$,
Now, we know that in general that, $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$ , where n is any real number.
So, $\dfrac{d}{dx}\left( \sqrt{x} \right)=\dfrac{1}{2}{{x}^{\dfrac{1}{2}-1}}$
Or, $\dfrac{d}{dx}\left( \sqrt{x} \right)=\dfrac{1}{2\sqrt{x}}$
So, $\dfrac{dy}{dx}=\dfrac{1}{2\sqrt{x}}$
Now, taking dx from denominator of left hand side to numerator of right side, we get
$dy=\dfrac{dx}{2\sqrt{x}}$
Putting x = 9 and dx = -2, we get
$dy=\dfrac{-2}{2\sqrt{9}}$
On solving, we get
$dy=-\dfrac{1}{3}$
We know that $\dfrac{1}{3}=0.333333.....$
So, $dy=-0.333$
As we discussed above that,
$\sqrt{x+\delta x}=y+dy$ ,
Putting values of x = 9, $\delta x=-2$ and dy = -0.333, we get
$\sqrt{9+(-2)}=\sqrt{9}+(-0.333)$
On simplifying, we get
$\sqrt{9+(-2)}=3+(-0.333)$
On solving, we get
$\sqrt{7}=2.667$
( ii ) Now, we will solve for finding square root of 2.5
Now, nearest number to 2.5 whose square root can be taken is 4, as $\sqrt{4}=2$ , so let us consider that x = 4 and $\delta x=dx=-1.5$ , as $x+\delta x=4+(-1.5)=2.5$
Now, the square root of x will lead to any number equals to say, y.
So, consider $y=\sqrt{x}$……( i );
As y is dependent on x so x is independent variable and y is dependent variable.
Differentiating ( i ) with respect to x, we get
$\dfrac{dy}{dx}=\dfrac{d}{dx}\left( \sqrt{x} \right)$,
Now, we know that in general that, $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$ , where n is any real number.
So, $\dfrac{d}{dx}\left( \sqrt{x} \right)=\dfrac{1}{2}{{x}^{\dfrac{1}{2}-1}}$
Or, $\dfrac{d}{dx}\left( \sqrt{x} \right)=\dfrac{1}{2\sqrt{x}}$
So, $\dfrac{dy}{dx}=\dfrac{1}{2\sqrt{x}}$
Now, taking dx from denominator of left hand side to numerator of right side, we get
$dy=\dfrac{dx}{2\sqrt{x}}$
Putting x = 4 and dx = -1.5, we get
$dy=\dfrac{-1.5}{2\sqrt{4}}$
On solving, we get
$dy=-\dfrac{1.5}{2.2}$
$dy=-0.375$
As we discussed above that,
$\sqrt{x+\delta x}=y+dy$ ,
Putting values of x = 4, $\delta x=-1.5$ and dy = -0.375, we get
$\sqrt{4+(-1.5)}=\sqrt{4}+(-0.375)$
On simplifying, we get
$\sqrt{2.5}=2+(-0.375)$
On solving, we get
$\sqrt{2.5}=1.625$

Hence, $\sqrt{7}=2.667$ and $\sqrt{2.5}=1.625$

Note: To solve such a question, this method must be known by students as this helps students in solving these questions easily and accurately. One must always know that $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$, where n is any real number. Try not to make any calculation mistakes.