Latest Elitmus Aptitude Question SOLUTION: Q. How many solutions exist for (x+5)^1/3=x((x+5)^1/3) Option a) 1 b) 2 c) 3 d) 4

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Q. How many solutions exist for (x+5)^1/3=x((x+5)^1/3)
Option
a) 1
b) 2
c) 3
d) 4

Asked In Elitmus vivek priyadarshi (10 years ago)

Unsolved Read Solution (7)

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Read Solution (7) :

RAKESH  (10 years ago)

(x+5)^1/3=x*[(x+5)^1/3]
or (x+5)^1/3-x*[(x+5)^1/3]=0
or (x+5)^1/3*(1-x)=0
or (x+5)^1/3=0 ,1-x=0
or x=-5 ,x=1
only two solution
option(2)

Is this solution Helpfull? Yes (44) | No (1)  

julee verma  (10 years ago)

option 2)-
(x+5)^1/3-x(x+5)^1/3=0
(x+5)^1/3[1-x]=0
x=-5,1 means two solution

Is this solution Helpfull? Yes (6) | No   

Madhusudhan  (10 years ago)

Ans b) 2

we can solve it in two ways:

ONE:

(x+5)^1/3=x((x+5)^1/3)

(x+5)^1/3 - x((x+5)^1/3) = 0

(x+5)^1/3(1-x) = 0

(x+5)^1/3 = 0 or (1-x) = 0

x = -5 or x = 1

TWO:

By Cubing both sides we get,

(x+5)^1/3=x((x+5)^1/3)

(x+5) = x^3 (x+5)

By solving it, we get
equation:
x^4 + 5x^3 - x - 5 = 0

x^3 (x + 5) - 1(x + 5) = 0

(x^3 - 1)(x + 5) = 0

x^3 - 1 =0 or x = -5

x^3 = 1 => If it is square we can assume that, there would be two solutions +1 ,-1
for odd powers, sign is same as before.

Is this solution Helpfull? Yes (5) | No (1)  

kulkarni  (10 years ago)

rise to 3rd power
(x+5)=x^3((x+5))
degree=4
so 4 solutions

Is this solution Helpfull? Yes (1) | No (3)  

vikas kumar  (10 years ago)

here (x+5)^1/3=x((x+5)1/3)
x+5=x^3(x+5) {cubic on both side}
x^3(x+5)-(x+5)=0
(x+5)(x^3-1)=0
now, x+5=0,x^3-1=0
x=-5, (x-1)(x^2+x+1)=0
x-1=0,x^2+x+1=0
x=1,and x^2+x+1=0 have two roots
x1=(-1+3^1/2i)/2
x2=(-1-3^1/2)/2
so total roots is -5,1,x1 and x2 ans=4

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Ajoy Singh  (10 years ago)

4
sol:>
cube both side we get
(x+5)=x^3(x+5)
=>(x+5)(x^3-1)=0
x=-5,1,w,w^2 where
w=(-1+(3^1/2)i)/2,w^2=(-1-(3^1/2)i)/2 i=(-1)^1/2

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Himanshu Srivastava  (9 years ago)

this equality will be true only for x=1
for x=-5
it will cause (x+5)^1/3/(x+5)^1/3=0/0 form
which is indefinite form so only x=1

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