How do you find the square root of 1242?

3 Answers
Jun 7, 2017

Answer:

It's #3*sqrt138#

Explanation:

As we can see, 1242 is not a perfect square, and so therefore, will not be able to be simplified into a whole number. What we can do, though, is simplify #sqrt1242# until it's in the form #a*sqrtb#, where #sqrtb# cannot be simplified further.

Let's start by dividing #sqrt1242# into its factors:

#sqrt1242=sqrt2*sqrt621=sqrt2*sqrt3*sqrt207=sqrt2*sqrt3*sqrt3*sqrt69=sqrt2*sqrt3*sqrt3*sqrt3*sqrt23#

That last expression can be rewritten as:

#sqrt2*sqrt3*sqrt9*sqrt23#

And that can be rewritten as:

#3*sqrt2*sqrt3*sqrt23#

Further simplification:

#3*sqrt138#

Nov 8, 2017

Answer:

#sqrt(1242) ~~ 12171889/345380 ~~ 35.2420204#

Explanation:

To find rational approximations to #sqrt(1242)# we can proceed as follows:

First split #1242# into pairs of digits starting from the right:

#12"|"42#

Examining the leftmost group of digits, note that:

#3^2 = 9 < 12 < 16 = 4^2#

So:

#3 < sqrt(12) < 4#

and:

#30 < sqrt(1242) < 40#

In fact note that #12# is close to the average of #9# and #16#, so a reasonable approximation to #sqrt(12)# is #3.5# and to #sqrt(1242)# is #35#.

Let's use a variant on the Babylonian method:

Given a rational approximation #p_i/q_i# to #sqrt(n)#, a better approximation is given by #p_(i+1)/q_(i+1)# where:

#{ (p_(i+1) = p_i^2+n q_i^2), (q_(i+1) = 2 p_i q_i) :}#

Starting with #p_0/q_0# repeatedly apply these formulae to get better approximations.

Let #n=1242# and #p_0/q_0 = 35/1#

Then:

#{ (p_1 = p_0^2+n q_0^2 = 35^2+1242 * 1^2 = 1225+1242 = 2467), (q_1 = 2 p_1 q_1 = 2 * 35 * 1 = 70) :}#

#{ (p_2 = p_1^2+n q_1^2 = 2467^2 + 1242 * 70^2 = 6086089 + 6085800 = 12171889), (q_2 = 2 p_1 q_1 = 2 * 2467 * 70 = 345380) :}#

So:

#sqrt(1242) ~~ 12171889/345380 ~~ 35.2420204#

Dec 21, 2017

Answer:

35.2420...

Explanation:

#sqrt(12'42.00'00'00'00')=35.2420#
#-3^2#
#"------------"#
#color(white)(...)342color(white)(..........................)65xx5=325#
#-325#
#"------------"#
#color(white)(.....)1700 color(white)(........................)702xx2=1404#
# color(white)(.)-1404#
#"------------"#
#color(white)(........)29600 color(white)(............. ......)7044xx4=28176#
# color(white)(....)-28176#
#"--------------"#
#color(white)(..........)142400 color(white)(..................)70482xx2=140964#
# color(white)(.....)-140964#
#"--------------------"#
#color(white)(..............)143600color(white)(...............) 704840xx0=0#