How you elementary find integer part of #(7,2-sqrt10)/(3,5+0,5)#?

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Answer 1 · 2017-06-23T18:35:50

#1#

Let #m# be the integer part of #(7,2-sqrt10)/(3,5+0,5)# then

#4.0 m = 7.2-sqrt 10# or

#10 = (7.2-4.0 m)^2 rArr 16m^2-2*4.0*7.2m+7.2^2-10=0#

now solving for #m#

#m = 1.00943# and #m = 2.5906#

and we follow with #m = 1#

Answer 2 · 2017-06-23T22:37:55

#1#

Note that:

#3.0^2 = 9 < 10 < 10.24 = 3.2^2#

Hence:

#3.0 < sqrt(10) < 3.2#

Multiply all parts by #-1# and reverse the inequalities to get:

#-3.2 < -sqrt(10) < -3.0#

Add #7.2# to all parts to get:

#4.0 = 7.2-3.2 < 7.2-sqrt(10) < 7.2-3 = 4.2#

Also:

#3.5+0.5 = 4.0#

Hence:

#1.0 = 4.0/4.0 < (7.2-sqrt(10))/(3.5+0.5) < 4.2/4.0 = 1.05#

So the integer part of #(7.2-sqrt(10))/(3.5+0.5)# is #1#