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by kvcpk » Wed Jul 14, 2010 9:52 am
pullagurla wrote:Extract the square root of [13/4]^4 - [13/3]^4 / [13/4]^2 - [13/3]^2
a)65/12
b)77/12
c)78/12
d)113/12
e)none
[13/4]^4 - [13/3]^4 / [13/4]^2 - [13/3]^2
=[13/4]^4 - [13/3]^4 * ([13/4]^2 + [13/3]^2) / [13/4]^4 - [13/3]^4 [Multiply and div by ([13/4]^2 + [13/3]^2) ]
=([13/4]^2 + [13/3]^2)
=13^2 (1/16 + 1/9)
=13^2(25/16*9)
sqrt= 13(5/3*4)=65/12
pick A
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by indiantiger » Fri Jul 16, 2010 8:21 am
as the numbers involve fraction and powers, what I do is substitute alphabets in place of the fractions or large numbers so I dont get confused:

lets say 13/4 = a
and 13/3 = b

which makes the equation =( a^4-b^4 )/( a^2 -b^2)
=( a^2+b^2 )* (a^2-b^2) / a^2 - b^2
= a^2+b^2
now substitute the values of a and b
=( 13/4 ) ^2 + (13/3) ^2
= 13*13/4*4 + 13*13/3*3
take 13^2 common
= 13^2 ( 1/ 4^2 + 1/3^2)
=sqrt (13^2/4^2*3^2) * 25
= 13*5/4*3
= 65/12 (answer)
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