can anyone help me to find the square root of non perfect squares?

Question

Can anyone give me good and more easier method to find the square roots of non perfect squares .It finds very hard for me to calculate without calculator in my exams

Answer 1

It reminds me of a small trick when I first learned calculus:

$$\frac{\sqrt{x+h}-\sqrt{x}}h\approx \frac1{2\sqrt x}$$ for small $h$, due to the first principle of calculus.

By rearranging, $$\sqrt{x+h}\approx\sqrt x+\frac{h}{2\sqrt x}$$

For example, $$\sqrt{4.1}=\sqrt{4+0.1}\approx\sqrt4+\frac{0.1}{2\sqrt4}=2.025$$

which has a percentage error of $0.0076$%.

---

Indeed, there is a step by step procedure to take square root by hand.

Given a number $x$, estimate the number of digits of $\sqrt x$ (this is not so hard for small $x$).

Then, let $\sqrt x=\overline{ABC\cdots.abc\cdots}$ (that’s a decimal dot).

Due to my poor english, I would instead show a worked example.

Compute $\sqrt{123}$ rounded down to the nearest tenth.

Let $\sqrt{123}=\overline{AB.a}+\cdots$

1. Find the largest integer $A$ such that $(10A)^2\le 123$

We get $A=1$.

2. Find the largest integer $B$ such that $(10A+B)^2\le 123$

We get $B=1$.

3. Find the largest integer $a$ such that $\left(10A+B+\frac{a}{10}\right)^2\le123$

We get $a=0$.

So $\sqrt{123}$ rounds down to $11.0$.

Answer 2

one can do Newton's method by hand. Some judgment is required in the number of digits in the division step.

For $\sqrt {61}$ the estimate $8$ is probably closer than $7$

$$ \frac{61}{8} = 7.625 $$ $$ \frac{8 + 7.625}{2} = 7.8125 $$ $$ $$ $$ $$ $$ \frac{61}{7.8125} = 7.808 $$ $$ \frac{7.8125 + 7.808}{2} = 7.81025 $$ $$ $$ $$ $$ $$ \frac{61}{7.81025} \approx 7.810249352 $$ $$ \frac{7.81025 + 7.810249352}{2} \approx 7.810249675 $$ $$ $$ $$ $$

The attractive aspect is that the number of correct digits roughly doubles ate each step $$8$$ $$7.8125$$ $$7.81025$$ $$7.810249675906661$$ $$7.810249675906654$$

Answer 3

If you're happy with fractions, I like the "Babylonian method" (you can Google it).

To estimate $\sqrt{k}$, start with a guess $x_0$ then use the iteration $x_{n+1}=\frac{1}{2}(x_{n}+\frac{k}{x_{n}})$.

For example, estimating $\sqrt{5}$ : $$\begin{align} x_0&=2 \\ x_1 &=\frac{1}{2}(2+\frac{5}{2}) = \frac{9}{4} \\ x_2 &=\frac{1}{2}(\frac{9}{4}+\frac{5}{\frac{9}{4}}) =\frac{1}{2}(\frac{9}{4}+\frac{20}{9}) = \frac{161}{72} \\ & \mathrm{etc}. \end{align}$$

Answer 4

If you want $\sqrt{N}=\sqrt{a^2+b}\quad$ with $a$ being the greatest integer possible such that $b\ge 0$.

Then by factoring $a$ you get by Taylor expansion: $$x_0=a\left(1+\frac b{a^2}\right)^\frac 12\approx a\left(1+\frac b{2a^2}\right)\approx a+\frac b{2a}$$

Starting from this seed you can apply the Newton method to $f(x)=x^2-N$.

This gives $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$

In particular $\require{cancel}f(x_0)=(a+\frac b{2a})^2-(a^2+b)=\cancel{a^2}+\cancel{2a\frac b{2a}}+(\frac b{2a})^2-\cancel{a^2}-\cancel{b}$

Thus a good approximation after just $1$ iteration is given by

$$x_1=\left(a+\frac b{2a}\right)-\frac{\left(\frac b{2a}\right)^2}{2\left(a+\frac b{2a}\right)}$$

Once simplified it can be written: $$\dfrac{8a^4+8ba^2+b^2}{4a(2a^2+b)}$$

This generally is enough to have a few significant digits right:

$\sqrt{5}=\sqrt{2^2+1}\approx\dfrac{161}{72}\approx 2.236111\quad[2.236067]$

$\sqrt{91}=\sqrt{9^2+10}\approx\dfrac{59068}{6192}\approx 9.539405\quad[9.539392]$

I think the advantage of this method with $a,b$ is that the formula in yellow is not too difficult to remember for your examination and it gives you a fair result already.

Answer 5

Method 1: The Generalized Binomial Theorem (Newton): If $|x|<1$ then $(1+x)^r=1+A_1x+A_2x^2+A_3x^3+...$ where $A_1=r$ and $A_{n+1} =A_n\cdot \frac {(r-n)}{n}.$ For example $\sqrt {61}=8(1+r)^{1/2}$ with $r=-3/64.$

Method 2: About 19 centuries ago Heron (Hero) of Alexandria wrote of letting $x_{n+1}=x_n-\frac {x_n^2-A}{2x_n},$ when $A>0$ and $x_1>0,$ to make a sequence $(x_1,x_2,x_3,...)$ converging to $\sqrt A.$ This is a special instance of Newton's Method, as already covered in another answer. A "tweak" to this is to, instead, let $x_{n+1}=x_n-\frac {2x(x_n^2-A)}{3x_n^2+A}$, in which the number of decimal places of accuracy approximately triples at each iteration.