Latest Elitmus Aptitude Question SOLUTION: how many solutions exist for sqrt(x+5)=x(sqrt(x+5)) options are a)1 b)2 c)3 d)4

• Placement Papers
• /
• Elitmus
• /
• Numerical Ability
• /
• Probability

Elitmus Question Papers:

(#M40038672) ELITMUS Numerical Ability question Number System Keep an EYE

how many solutions exist for sqrt(x+5)=x(sqrt(x+5))
options are a)1 b)2 c)3 d)4

Asked In Elitmus Amit Kumar Pal (9 years ago)

Unsolved Read Solution (17)

Is this Puzzle helpful?   (6)   (2) Submit Your Solution Probability

Read Solution (17) :

vikash kumar  (9 years ago)

x(sqrt(x+5))-sqrt(x+5)=0
(x-1)(sqrt(x+5)=0
so,
x-1=0 or x+5=0;
so,
x=1 or -5

Is this solution Helpfull? Yes (31) | No (3)  

akash verma  (9 years ago)

ans: c i.e 3
(x+5)^1/2 = x(x+5)^1/2
squaring both side
then we get
x+5 = x^2(x+5)
x+5 - x^2(x+5) = 0
take x+5 as common
x+5(1-x^2)=0
(x+5)(1+x)(1-x)=0 i.e a^2-b^2
x=-5;
x=-1;
x=1

Is this solution Helpfull? Yes (18) | No (16)  

chandan  (9 years ago)

x=-5,-1,&1
but
x=-1 does not satisfy the eqn.
so only 2 are ans
(b)

Is this solution Helpfull? Yes (14) | No   

Saikat Guin  (9 years ago)

after calculation we get x^2=1 so, x=1 or -1. so, 2 solutions exist. ans: (b)

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

sakshi singh  (9 years ago)

ans is c ie 3

Is this solution Helpfull? Yes (2) | No (7)  

Divya  (9 years ago)

Why -1 cannot be considered? if we put x=-1 in the equation we get:
sqrt(-1+5)=-1(sqrt(-1+5))
=>sqrt(4)=-1(sqrt(4))
=>(+-)2=-1[(+-)2]
=>(+-)2=(-+)2
L.H.S.=R.H.S
Why this caanot be considered?

Is this solution Helpfull? Yes (1) | No (4)  

DILIP KUMAR YADAV  (9 years ago)

sqrt(x+5)=x sqrt(x+5)
square both side,
x+5=x^2 (x+5)
x^3+5x^2+x+5=0
(x^2+1)(x+5)=0
x=1,-5
x=1

Is this solution Helpfull? Yes (1) | No   

Vaishak Damodar  (9 years ago)

Come on guys!!! Ans is clearly 3. Let m explain in detail :
By squaring on both sides you get : x+5 = x^2(x+5)
==> x+5 = x^3 + 5(x^2)

==> (x^3) + 5(x^2) - x - 5 = 0
If you solve the above equation, we get 3 values for x i.e : +1, -1, -5 . Therefore, 3 solutions. Substitute them in the above equation, you will get the answer.

Is this solution Helpfull? Yes (1) | No (2)  

Tuhina Bhattacharya  (5 years ago)

Given sqrt(x+5)=x(sqrt(x+5)),
Squaring both the sides, we get: (x+5)=x^2(x+5)
Cancelling the equal terms i.e. (x+5) from both sides, we get: 1=x^2
Now, taking the square root on both sides, we get
1=x;
which means either +1 or -1 can give 1,
Thus it has 2 roots.

Therefore , Answer is 2.
Option B is the Answer.

Is this solution Helpfull? Yes (1) | No   

Amit kumar singh  (9 years ago)

ans will be 3 soluton
1st direcrtly cancel sqrt(x+5) frem both side we get x=1
then bring above equestn in polynomial frm by squaring bth side
then divide it by x-1 to get further roots of x

Is this solution Helpfull? Yes | No (2)  

Pranoy Bej  (9 years ago)

ANSWER==> b)2
x=1/-5

Is this solution Helpfull? Yes | No   

Mohit Mathuriya  (9 years ago)

ans is 3
suaring both side..x=5,-1,1

Is this solution Helpfull? Yes | No (3)  

SUDHAKAR SRIVASTAVA  (9 years ago)

c-3 solutions

Is this solution Helpfull? Yes | No (2)  

shabana ismail  (9 years ago)

sq(x+5)=xsq(x+5) : squaring x+5 = x^2(x+5)
x+5/(x+5)=X^2 , so x = +-1, so solution 2,,,

Is this solution Helpfull? Yes | No (1)  

ASHISH MUKHERJEE  (9 years ago)

we got x= -5,-1 and 1

but put -1 in original eq:- sqrt(-1+5) =-1(sqrt(-1+5))
-> 2 = -2
so,-1 not satisfied the above equation so only two solution can be possible.
Ans: b)2

Is this solution Helpfull? Yes | No   

abhin  (9 years ago)

sqrt(x+5)=x(sqrt(x+5))
squaring both sides
(x+5)=x^2(x+5)
x^2=1=>x=+/- 1
so nly 2 solution

Is this solution Helpfull? Yes | No   

sujeet kumar  (5 years ago)

sqrt(x+5)=x(sqrt(x+5))
Squring both side
x+5=x^3+5x
=> x^3+4x+-5=0
=>(x-1)(x^2+x+5)=0
x=1 is a solution
and (x^2+x+5)=0
D=b^2-4ac
D=1-20=-19
so there is no real solution apart from 1
Correct answer is a) 1

Is this solution Helpfull? Yes | No   

Please Login to Read Solution.