The real roots of the equation r2/3...
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The real roots of the equation r2/3+x/3 - 2 = 0) are (a) 1,8 (b) -1,-8 (c) 1,8 (d) 1,-8

JEE/Engineering Exams
Maths
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The given equation is ( x^{2 / 3}+x^{1 / 3}-2=0 ) Put ( x^{1 / 3}=y, ) then ( y^{2}+y-2=0 ) ( Rightarrow(y-1)(y+2)=0 ) ( Rightarrow y=1 ) or ( y=-2 ) ( Rightarrow x^{1 / 3}=1 quad ) or ( quad x^{1 / 3}=-2 ) ( therefore x=(1)^{3} ) or ( x=(-2)^{3}=-8 ) Hence, the real roots of the given equations are 1,-8
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