IIT JEE , AIEEE Forum - Mathematics , Algebra

bachinarayudu

sum of three numbers is 6,sum of their squares is 8, and sum of their fourth powers is?

hash_include

i'm afraid thats not possible mate.

we know that

$\frac {a^2 + b^2 + c^2}{3} \ge (\frac {a + b + c}{3})^2$

$=>\frac {a^2 + b^2 + c^2}{3} \ge 4$

$=> a^2 + b^2 + c^2 \ge 12$

which is not the case in the given sum.

an argument can be given for negative numbers also. because, if any one of a,b or c is negative, the sum of their squares will obviously be more than if they were positive
Smile

pramod6990

rightly said maaeen.....

not possible yaar......

hsbhatt

didnt say the numbers are all real did he?

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