(1) (x+2)²+(x+3)³+(x+4)⁴=0, solve this equation

Are you sure of the question? The original problem had 2 on the RHS. That makes it easier to solve.

I AGREE WITH U SIR.

NOW MY SOL. BY THIS ASSUMPTION IS

A.M OF (X+3) AND (X+4)=

 

(X+3+X+4)/2=X+3



 TAKING X+3=Y

 

IT IS

 

(Y-1)^2 +Y^3 +(Y+1)^4=2

Y^2-2Y+1 +Y^3 +Y^4+ 4C1 (Y^3) +4C2(Y^2) +4C3(Y) +1=2      (EXPANDING (Y+1)^4 BINOMAL EXPANSION)

IT BECOMES

 Y^4+5Y^3+7Y^2+2Y=0

SOL. =>Y=0



 

THEREFORE x+3=Y

X=-3

RATE ME IF USEFUL.

But, there are 3 more roots awaiting discovery.

OK Y = -2 IS ANOTHER ROOT

THEREFORE X=-2-3

X=-5

NEXT TWO ARE ALSO FOUND.

DIVIDING THE POLYNOMIAL WITH (Y+2)

WE GET (Y+2)(Y^3+3Y^2+Y)=0

THAT GIVES Y=0 AND Y=[-3 +-(ROOT) 5] /2

NOW PUT THEM IN EQ. AND GET X