Solve x^(2/3)-2*x^(1/3)=15

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14 Jun 2013 Solve

Solve
x^(2/3)-2*x^(1/3)=15

OPtion

1) x=1,x=8
2) x=4,x=-64
3) x=3,x=-27
4) x=5,x=125
5) x=27,x=-64
6) x=125,x=64
7) x=125,x=-27
8) x=9,x=125
9) x=7,x=144
10) None of these

Solution

Put x^(1/3)=t to get t^2-2t=15
solve
t=5,t=-3
therefore x^(1/3)=5, x^(1/3)=-3
Taking cube on both sides, we get
x=125,x=-27
option 7

Correct Option: 7 Best Solution (22)

aman tiwari   11 years AGO

let x^(1/3)=a
a^2-2*a-15=0
a^2-(5-3)a-15=0
a^2-5a+3a-15=0
a(a-5)+3(a-5)=0
(a-5)(a+3)=0
putting the value of a
x^(1/3)=5
i.e x=125
x^(1/3)=-3
i.e x=-27

Like? Yes (11) | No (4)  

Harish yadav   11 years AGO

Let x^(1/3)=a...thdn given eqtn reduce to a^2-2a-15=0....by solving we get a=-3 0r a=5....replace the value of a & cubing both sidethen we get x=125 & x=-27

Like? Yes (1) | No   

saisaranya   11 years AGO

let c=x^1/3 and the eqn is c^2-2c-15 solving we get c=-3 and c=5 then x=125 and x=-27

Like? Yes (1) | No   

Karan   11 years AGO

x^(2/3)-2*x^(1/3)=15
x^(2/3)-2*x^(1/3)-15 =0
[(x^(1/3) -5]*[(x^(1/3) +3]=0
when [(x^(1/3) -5]= 0
[(x^(1/3)=5]
x=5^3=125
when
[(x^(1/3) +3]=0
[(x^(1/3)] = -3
x=-27

Like? Yes (4) | No (5)  

Adarsha Upadhya   11 years AGO

x^1/3.x^1/3-2.x^1/3-15=0
let a=x^1/3;
a^2-2a-15=0;
a=-3 and a=5;
-3=x^1/3 on solving x=-27
5=x^1/5 on solving x=125

Like? Yes (3) | No (5)  

dev kumar   11 years AGO

let
k=x^1/3
eqn will be---:
k^2-2k-15=0
solve this quardratic and then
k=5,k=-3
but
k=x^1/3
so
x=125&-27

Like? Yes (3) | No (6)  

C.Manohar   11 years AGO

x^(1/3)=y
y^2-2y-15=0
(y-5)(y+3)=0
y=5=>x=5^3=125.
y=-3=>x=(-3)^3=-27.

Like? Yes (3) | No (6)  

krishnapriya   11 years AGO

given the eqn. x^(2/3)-2*x^(1/3)=15
let x^(1/3)=u
therefore, u^2-2u-15=0
solving the above we get u=5 (or) u=-3
(ie)x^(1/3)=5 (or) x^(1/3)=-3
cubing both sides we get
x=125 (or) x=-27

Like? Yes (2) | No (5)  

pradeep kumar korikana   11 years AGO

let x^1/3 =y
sub in given expression
y^2-2y=15
Y(y-5)+3(y-5)=15
y=-3 or y = 5
again sub in x^1/3 =y
we will get x=-27 or x=125

Like? Yes (1) | No (4)  

shailesh kumar mishra   11 years AGO

put u = x^(1/3 then we get u square-2u-15=0
by solving we get u=5 and u=-3
so 5=x quibik root the x=125
and -3= x quibik root then x=-27

Like? Yes (1) | No (4)  

Jigyasa Kumar   11 years AGO

125 and -27 satisfies the equation .

Like? Yes (1) | No (4)  

Aswini Kumar Nayak   11 years AGO

x^(2/3)-2*x^(1/3)=15
after rearrange

(x^(1/3))^2-2*1*x^(1/3)+1=16
(x^(1/3)-1)^2=16
(x^(1/3)-1)=+4 or -4
then value of x=125 or -27

Like? Yes (1) | No (4)  

IFFAT IRAM   11 years AGO

let x^(1/3) be Z
Z^2 - 2Z - 15 = 0
solving,
Z=5, -3
x=125, x= -27

Like? Yes (2) | No (5)  

Sneha Agarwal   11 years AGO

let x^1/3 = y
then
y^2 - 2y-15=0
solving we get y= 5, -3
x^1/3 = 5, -3
x = 125, -27

Like? Yes (1) | No (4)  

Balakrishnan M   11 years AGO

x^(2/3)-2*x^(1/3)=15
x^(2/3)-2*x^(1/3)-15=0
x^(2/3)-5*x^(1/3)+3*x^(1/3)-15=0
(x^(1/3)-5)(x^(1/3)+3)=0
(x^(1/3)-5)=0 or (x^(1/3)+3)=0
x^(1/3)= 5 or x^(1/3)= -3
x=5^3 or x=(-3)^3
x=125 or x=-27

Like? Yes (3) | No (6)  

Browns MN   11 years AGO

Let x^(1/3)= y--->1
y^2 - 2y - 15 =0
Solving y=5,y=-3
From 1, x=125 and x= -27

Like? Yes (1) | No (4)  

Anbalagan K   11 years AGO

Let x^1/3=a
a^2-2a=15
a^2-2a-15=0
Solving the equation, we get a=5 or -3
x=a^3 i.e., 125 or -27

Like? Yes (1) | No (4)  

vinod   11 years AGO

verify options

Like? Yes (2) | No (5)  

ravi singh   11 years AGO

let y= x^(1/3)

then equation will given as
y^2 -2y =15

=> (y-1)^2 = 16
=>y= -3, 5

therefore x= -27, 125

Like? Yes (1) | No (4)  

tanvi agarwal   11 years AGO

let x^1/3=p
eqution
p^2-2p-15=0
p=5,-3
hence
x=125
x=-27

Like? Yes (1) | No (4)  

Devendra Marghade   11 years AGO

x^(2/3)-2*x^(1/3)=15
If x^1/3=a, above eqn. becomes, a^2-2*a-15=0
or (a-5)(a+3)=0, a=5,-3
So x=a^3=5^3=125 & -3^3=-27

Like? Yes (1) | No (5)  

deep   11 years AGO

let;
x^1/3=y;
then
y^2-2y=15
hence, y=-3,5

Like? Yes (1) | No (5)