An Important mathematics for students!

Hello my friends and students,
hope you are all well.
In this video we are going to solve a nice exponential equation for the students of class x.
Please watch the full video and if you like how to solve this problem please like , share, comment and subscribe ❤️🙏
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#exponentialequation
#mathstricks
#exponentialformula
#algebraequation
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[@annabanerjee5164]

Verified, The roots satisfied with the original equation. Thank you for solution!

[@DebdasBandyopadhyay-yq5jg]

Thanks for watching, welcome

[@daviddoby9648]

Équation (1): X^(2/3)+ X^(1/3)=2 Let Y=X^(1/3)=³√X (1) : [X^(1/3)]^2 + [X^(1/3)]=2 (1) Becomes : (2) : Y^2 + Y -2 =0 Équation of the 2nd degree that we can solve easily : 1 is évident solution of (2) So (2) is équivalent with (Y-1)(Y+2)=0 Two solutions, Y=-2 or Y=1 Or, using Delta d=b²-4ac= 1²-4*1*(-2)=1+8=9 d >0 so 2 solutions Y1=(-b-√d)/(2a)=(-1-√9)/2=(-1-3)/2=-4/2=-2. Y2=(-b+√d)/(2a)=(-1+√9)/2=(-1+3)/2=2/2=1. ³√X=Y so X=Y³ X1=Y1^3=(-2)^3=-8 X2=Y2^3=1^3=1 Solution in R={1 ; -8} So thé answer is X= {1 or -8}

[@DebdasBandyopadhyay-yq5jg]

Nice