x^2-8/x plzz solve it fast
Anonymous:
X^2-8/x = 0 ??
is the given equation = 0
no
u hv to factorise it
dear solving means just like roots finding so , with equate to zero you can't find solution if questions ask factorization then no required to equate to zero
yes
Answers
Answered by
5
x^2 - 8/x
=(x^3 -8)/x
=(x^3 - 2^3)/x
=(x-2)(x^2+2x+4)/x
Since
(a3-b^3)=(a-b)(a^2+ab+b^2)
=(x^3 -8)/x
=(x^3 - 2^3)/x
=(x-2)(x^2+2x+4)/x
Since
(a3-b^3)=(a-b)(a^2+ab+b^2)
my answer is same as u posted hrik
but u have to ask like this
factorize the expression
no 1/x is missing
solve is different from above
okk
u see 1/x is missing
1/x is nothing but denominator is x
in my answer it is there check again
Thank you selecting as brainliest
Answered by
4
x² -8/x
take LCM
(x³ -8)/x
(x³ -2³)/x
use a³ - b³ = (a -b)( a² + ab + b²)
(x -2)(x²+2x +4)/x
now equate to zero for finding roots
(x -2)(x² +2x +4)/x =0
x ≠0
so,
x =2 and x² +2x +4 =0 it has no real roots bcoz discriminant = 2² -4×4 < 0
so, x = 2 is answer
take LCM
(x³ -8)/x
(x³ -2³)/x
use a³ - b³ = (a -b)( a² + ab + b²)
(x -2)(x²+2x +4)/x
now equate to zero for finding roots
(x -2)(x² +2x +4)/x =0
x ≠0
so,
x =2 and x² +2x +4 =0 it has no real roots bcoz discriminant = 2² -4×4 < 0
so, x = 2 is answer
but answer is
1/x (x-2) (x^2+2x+4)
okk abhi and mystcd
u sincerely see the answer , here all data given by sir and me
u'r solution also correct
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