rootover6+rootover6+rootover6....
Himanshu1629:
why is this question reported
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let x=√6+√6+√6+......
x= √6 + x(√x also)
square them
x²= 6 + x
x²-x-6=0
x²-3x+2x-6=0
x(x-3)+2(x-3)=0
(x+2)(x-3)=0
x=-2; x=3
x can't be -2 as it is under root
therefore, x=3
REGARDS
Himanshu1629
<Virtuoso>
x= √6 + x(√x also)
square them
x²= 6 + x
x²-x-6=0
x²-3x+2x-6=0
x(x-3)+2(x-3)=0
(x+2)(x-3)=0
x=-2; x=3
x can't be -2 as it is under root
therefore, x=3
REGARDS
Himanshu1629
<Virtuoso>
hope it helped
please mark brainliest if it really helped
correct solution nice
plz mark brainliest
how?
why is this question reported
dont know
oki oki sorry.... so you are NOT the guy who asked this question
no
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