Solve 4*Sqrt[x/(x+1)] + 13*Sqrt[(x+1)/x)] = 28

Dr.Dipin Singh   10 years AGO

4*Sqrt[x/(x+1)] + 13*Sqrt[(x+1)/x)] = 28
let
Sqrt[x/(x+1)] =y
then eqn becomes
4y+13/y=28
solving, we get
y= -13/2 , 1/2 = Sqrt[x/(x+1)]
hence x= -169/165, 1/3

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Surya Narayana K   10 years AGO

16*(x/x+1) + 169*(x+1/x) + 104 = 784
After solving,
495x^2 + 342x - 169 = 0
x = -169/165 and 1/3

Like? Yes (2) | No (1)  

TUSHAR    10 years AGO

put Sqrt[x/x+1]=p,
equation become, 4p+13/p=28
the root of equation is p1=13/2,p2=1/2
i.e. x1=-169/165, x2=1/3

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vishnuvardhan   10 years AGO

let x+1/x=t
4/sqrt(t)+13*sqrt(t)=28
t=4,4/169
x=1/3,-169/165

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nikhil   10 years AGO

on solving above given equation we get x=1/3 and on hit and trial method we get -169/165

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Aakash Jain   10 years AGO

when we solve this equation, this equation is looks alike 495*X^2+342*X-169=0. when we solve this equation,the result is X=-169/165,1/3

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suryam surada   10 years AGO

4*Sqrt[x/(x+1)] + 13*Sqrt[(x+1)/x)] = 28

>>square operation both sides..then we will get
495*x^2+342*x-169=0;

>> if you substitute all the options..
4 th option satisfy the equation

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arnold   10 years AGO

square on both side
we get
16(x/x+1)+169(x+1/x)+104=784
16(x/x+1)+169(x+1/x)-680=0
take LCM
we get
495(x^2)+395(x^2)-169=0
calculate quadratic root of above equation
x= -169/16, 1/3

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VIKASH RAY   10 years AGO

The given equation is,
4*Sqrt[x/(x+1)] + 13*Sqrt[(x+1)/x)] = 28
Now,Let, [x/(x+1)]=y-----------------(1)
Then the given equation becomes,
4*sqrt(y)+ 13* sqrt(1/y) = 28
Now solving this equation we get,
y=1/2, 13/2
Now put the value of y in the equation number 1 we get,
x=1/3, -169/165
Hence option number 4 is correct.

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Ann Theressa   10 years AGO

put sqrt(x/(x+1)) = y

4y + 13/y = 28
4y^2 -28y + 13 = 0

y = 13/2 , 1/2

x/(x+1) = y^2
x/(x+1) = (13/2)^2
x/(x+1) = 169/4
x = -169/165

x/(x+1) = y^2
x/(x+1) = 1/2^2
x/(x+1) = 1/4
x = 1/3

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ARJYA HALDER   10 years AGO

Let Sqrt*[x/(x+1)]=a
So, the equation becomes 4a+(13/a)=28
which gives a=(1/2),(13/2)
which gives x=(1/3),(-169/165)

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The Trojan   10 years AGO

complex number

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Naveen Ganesh   10 years AGO

Let x/x+1 = a^2 then equation becomes 4a + 13/a = 28
=> 4a^2 - 28a + 13=0
=> a= 13/2 , a=1/2
Substituting both the values of a we get, x=-169/165 and x=1/3

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Rajasekhar Reddy   10 years AGO

let sqrt[x/x+1] = P

then it becomes

4P^2-28P+13 = 0

SOLVING WE GET

P = 13/2 AND 1/2

EQUATING SQRT[X/X+1] TO 13/2 AND 1/2 WE GET

X = -169/165 AND 1/3

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srija   10 years AGO

by simplifyng given quatr eqn we will get the followng eqn...
495x*2+342x-169=0
after solving the eqn we can get the roots -169/165 , 1/3

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gaurav bhatt   10 years AGO

Squaring on both sides:
(4 *sqrt(x/x+1)+(13*sqrt(x+1/x))2=(28)2
16(x/x+1)+169(x+1/x)+104=784
16(x/x+1)+169(x+1/x)=680
16x2+169(x+1)2=680(x(x+1))
495x2+342x-169=0;

Finding factors:
X=-169/165 and x=1/3

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Ragaven   10 years AGO

solving the quadratic equation we get the answer (-169/165,1/3)

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Mukesh Nayal   10 years AGO

Squaring on both sides,
(4 *sqrt(x/x+1)+13*sqrt(x+1/x))2=(28)2
16(x/x+1)+169(x+1/x)+104=784
16(x/x+1)+169(x+1/x)=680
Taking lcm on left hand sides,
16x2+169(x+1)2=680(x(x+1))
On solving ,
495x2+342x-169=0;
Factorizing using shridharacharya method,
X=-b+sqrt(b2-4ac)/2a
We got,
X=-169/165 and x=1/3

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sarvesh jhawar   10 years AGO

square both the side and then some manipulation
495*x*x+342*x-169=0
which give solution 4.

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Udhaya kumar   10 years AGO

let u=(x)/(x+1)
16u^2-680u+169=0
u=0.25,42.25
so x=-(169/165),(1/3)....

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Vivek Jain   10 years AGO

4t+13/t=28
where t=Sqrt[x/(x+1)]
t=13/2 and 1/2
so x=-169/165 and 1/3

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Saikat Guin   10 years AGO

4* sqrt[x/x+1]+13sqrt[x+1/x]= 28
=> 16[x/x+1]+ 169[x+1/x] +104= 28
=> 16x^2+169x^2+169+338x= 680x^2+680x
=> 495x^2+342x- 169=0
=> x={-342±√(342^2+ 4*495*169)}/2*495 \\\\ as x={-b±√(b^2-4*a*c)}/ 2a
=> x= (-342±672)/990
=> x= -169/165 or 1/3

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Rajat Goyal   10 years AGO

Best Option

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llamar   10 years AGO

take squrt[x/(x+1)]=p& do simplifications as follows
4p^2-28p+13=0 =>4p^2-2p-26p+13=0
at the end u will get x=-169/165,1/3

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RAKESH   10 years AGO

4*Sqrt[x/(x+1)] + 13*Sqrt[(x+1)/x)] = 28

Sqrt[x/(x+1)] = y then

4y + 13/y = 28 => y=1/2 , 13/2

or, Sqrt[x/(x+1)] = 1/2 , 13/2 => x= -169/165, 1/3

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shilpi   10 years AGO

put x/(x+1) = y
and solve the equation then

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Sivanjali   10 years AGO

After squaring on both sides the equation will be

16*(x/x+1) + 169*(x+1/x) + 104 = 784
16x^2 + 169(x+1)^2 + 104(x*(x+1))=784(x*(x+1))
495x^2 + 342x - 169 = 0
On solving we get
x = -169/165 and 1/3

Like? Yes (1) | No (1)  

J.DHANDAPANI   10 years AGO

Let Sqrt[x/(x+1)=t.
On sub in the given equation we get,
4t^2-28t+13=0.
on solving which gives t=6.5 , 0.5.
===>Sqrt[x/(x+1)]=6.5.
x/(x+1)=(6.5)^2
41.25 x=-42.25
x=-1.0242
=-169/165 and
===>Sqrt[x/(x+1)]=0.5.
x/(x+1)=(0.5)^2.
3x=1.
x=1/3.

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ADVAITA BASAK   10 years AGO

Solving the equation in general manner the answer become x=-169/165, 1/3

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Devendra Marghade   10 years AGO

4*Sqrt[x/(x+1)] + 13*Sqrt[(x+1)/x)] = 28
Substituting Sqrt[x/(x+1)]=t
4t+(13/t)=28
4t^2-28t+13=0, roots=13/2, 1/2
So Sqrt[x/(x+1)]= 13/2, x=-169/165 &
Sqrt[x/(x+1)]= 1/2, x=1/3

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PARVY GOVIL   10 years AGO

x=-169/165,1/3 i..e 4th option
reason :
let sqrt[x/x+1] = p
we have
4p + 13/p =28
4(p*p) -28p+13 =0
thus solving the above quadratic we get :
p=1/2;13/2
case 1:
p=1/2
1/2 = sqrt(x/x+1)
1/4=x/x+1
x=1/3
case 2:
sqrt(x/x+1)=13/2
x/x+1=169/4
x= -169/165
from case 1 & 2 we get
x= -169/165,1/3

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Kowshik Sarker   10 years AGO

let [x/(x+1)]=k^2
so [(x+1)/x]=1/k^2
putting this the equation becomes
(4*k)+(13/k)=28
which becomes 4*k^2-28k+13=0
solving this quadratic equation we get
k=13/2,1/2
so we get 2 equations to solve
1.x/(x+1)=(13/2)^2 and
2.x/(x+1)=(1/2)
solving this two equation we get the value of x which is x=-169/165 and x=1/3. So option 4 is correct.

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Soham Dalwadi   10 years AGO

suppose Sqrt[x/(x+1)]=a
then eq= 4a^2++13=28a
finding roots we get a=1/2,a=13/2
substituting a
4) x = -169/165, 1/3

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SUPROTIM PAUL   10 years AGO

SOLVING THE EQUATION WE GET THE ANSWER.ONE MAY LET THE SQRT PART AS 'A' FOR EASY TO CALCULATE

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megha   10 years AGO

Answer is 4) x=-169/165, 1/3

4*Sqrt[x/(x+1)] + 13*Sqrt[(x+1)/x)] = 28
By Squaring both sides, we get
(4*Sqrt[x/(x+1)] + 13*Sqrt[(x+1)/x)])^2 = 28^2
16*(x/x+1) + (169*(x+1)/x) + 2*4*13*1 = 784
((16x)/(x+1)) + 169 + (169/x) + 104 = 784
((16x)/(x+1)) + (169/x) = 511
16x^2 + 169x + 169 = 511(x^2 + x)
16x^2 + 169x + 169 = 511x^2 + 511x
495x^2 + 342x - 169 = 0
495x^2 + 507x - 165x - 169 = 0
3x(165x + 169) - 1(165x + 169) = 0
(165x + 169) (3x-1) = 0
(165x + 169)= 0
165x=-169, x = -169/165
3x-1=0, x = 1/3

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K.Bhagyasri   10 years AGO

by applying square on both sides
495X^2+342X-169=0

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Nitin Jaiswal   10 years AGO

expand the equation as 4x+13(x+1)=28*x^1/2*(x+1)^1/2
square both sides and find the root of the quadratic equation which is derived i.e,
495x62+342x-169=0
the roots are 1/3 and -169/165.

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