Finding roots of a complicated looking expression | Solve for x | PRMO RMO SSB IBPS

#ComplicatedExpression #SolveforX #SchoolMath #Algebra

[@dineshjangid3325]

Very nice explaination

[@AlfonsoNeilJimenezCasallas]

Awesome! 👏

[@teacher_charles]

Wonderful

[@Aparnafoodworks]

I have a doubt __BAA__ BBB__ AB__ 1)AABB 2)BBAB 3)BAAB 4)ABAB Which option is correct?

[@renuyadav8046]

Hi

[@renuyadav8046]

2

[@devondevon4366]

Answer x= 0, and x=-3 . got the same answer , however approach it differently. x^1/2 + x^-1/2/ 1- x + 1- x^-1/2/1+x^1/2 - (4+x)^1/2/1-x)^1/2 =0 My goal is to make all have a denominator of 1-x. So since the first fraction has x-1 it will be left alone. The second denominator and numerator will be multiplied by 1- x^1/2 to get 1- x^1/2 - x^-1/2 +1 The third denominator and numerator will be multiplied by (1-x)^1/2 to get (4+3x+x^2)^1/2 Therefore : x^1/2 + x^-1/2 1-x^1/2-x^-1/2+1 (4+ 3x + x^2)^1/2 = 0 --------------------- + ----------------------------- - --------------------------- 1-x 1-x 1-x 1+1 - (4+3x+x^2)^1/2 =0 ( x^1/2 and x^-1/2 canceled out between the first two fractions.) ------------------------------------------------------ 1-x 2 - (4+3x+x^2)^1/2 = 0 ( 1-x) (crossmultiply) 2 = (4+3x+x^2)^1/2 (add 4+3x+x^2)^1/2 to both sides) 4 = 4+3x+x^2 (square both sides) 0 = + 3x + x^2 (substrtact 4 from both sides) 0 = (3 +x)x (factor out x) 0 = x - 3 = x Answer x= -3 or 0 4:27

[@ayeshashaukat27]

Can we write a+b^3=(a+b)(a+b)(a+b)

[@mwhanmwhan8866]

Actually There are no solutions for this equation. Since x must differ from 0 because of x^(-1/2) . And x must be positif because it's under a square root

[@alfazero4906]

x doesn't have to be positive when it's under a square root. Sqrt(-3) = sqrt(3)*i, that's it. And yes, x must differ from 0 because we have x^(-1/2).

[@devondevon4366]

x= -3 and x=0