Solve 4{x^2+(1/x^2)} - 8{x+(1/x)} + 3 = 0

Akanksha Mishra   11 years AGO

Given
4{x^2+(1/x^2)} - 8{x+(1/x)} + 3 = 0

4x^2 + 4/x^2 - 8x - 8/x + 3 = 0
4x^4 - 8x^3 + 3x^2 + 4 = 0

put the both 2 and 1/2 you will get RHS Equals to zero

Like? Yes (15) | No (6)  

Anubala   11 years AGO

i did this puzzle by solving it as 4th power quadratic equation.

Like? Yes (1) | No   

P.Selva Dhasni   11 years AGO

the correct ans is 9)2,1/2.

according to BODMAS rule,

4{x^2+(1/x^2)}-8{x+(1/x)}+3=0

4((x^4+1)/x^2)-8((x^2+1)/x)+3=0

((4x^4+4)/x^2)-((8x^2+8)/x)+3=0

((4x^4+4)/x^2)-((8x^3+8x)/x^2)+3=0

((4x^4+4-8x^3-8x+3x^2)/x^2)=0

4x^4-8x^3+3x^2-8x+4=0

from the following equation the 'x' values are 2,1/2....

Like? Yes (1) | No   

Rituparna Chakraborty   11 years AGO

4{x^2+(1/x^2)}-8{x+(1/x)}+3=0
=>4{(x+(1/x)^2-2}-8{x+(1/x)}+3=0
=>4{x+(1/x)}^2-8-8(x+1/x)+3=0
put{x+(1/x)}=a so,
4a^2-8a-5=0
solution of this above equation,a=5/2,-1/2
so,x+(1/x)=-1/2=>2x^2+x+1=0,it's not possible.
so,x+(1/x)=5/2
=>2x^2-5x+2=0
So,x=2,1/2

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akansha   11 years AGO

consider x+(1/x)=y
solve the 2 quadratic equations
substitute the value of y as 1 +(1/x) and again solving the two equations we get,
x = 1/2,2

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vishnuvardhan   11 years AGO

put x+1/x=t
4(t^2-2)-8t+3=0
t=5/2,-1/2
x=2,1/2

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PRAGATHI G   11 years AGO

by trial and error method,pit -3,-2,-1,0,1,2,3 and see wic satisfies the given equatn,here 2 satisfies.. this implies(x-2) is a factor of given equatn.. divide given equatn by (x-2)we get a cubic equatn.. in this equation, we see tat 1/2 satisfies.. hence answers re 2,1/2

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ekta rani   11 years AGO

4{x^2+(1/x^2)}-8{x+(1/x)}+3=0
4{(x^4+1)/x^2}-8{(x^2+1)/x}+3=0
4(x^4+1)/x^2-8(x^2+1)/x +3=0
4(x^4+1)-8x(x^2+1)+3*x^2=0
4x^4+4-8*x^3-8x+3*x^2=0
4x^4-8*x^3+3*x^2-8*x+4=0
it gives 0 when x=2 and 1/2
this two values satisfy the equations
so 9 option is the solution of given equation

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sruthi   11 years AGO

answer is 2,1/2 because when substituting 2 and 1/2 the equation will be equal to 0.

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Rajasekhar Reddy   11 years AGO

let x+1/x = p
then the eqn becomes
4p^2-8p-5 = 0
by solving we get p = -1/2 and 5/2
for p = 5/2
x = 2,1/2

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utkarsh goel   11 years AGO

let x+1/x =p
then solve it we get p=-1/2,5/2
then using it we get the equation 2x^2-5x+2=0
and we get the values of x

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R.RAVIKUMAR   11 years AGO

Let x + 1/x = P
so, x^2 + 1/x^2 = P^2 -2
so, the equation will be.. 4p^2 - 8p -5 =0
by solving, we get p=5/2 and p=-1/2

with considering p=5/2..
by solving for x, we get x= 2 and 1/2

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Jaydeep Gupta   11 years AGO

2 nd 1/2 satisfies the given eqn

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Animesh Pal   11 years AGO

4{x^2+(1/x^2)}-8{x+(1/x)}+3=0
or, 4{x+(1/x)}^2-(10-2){x+(1/x)}-5=0
So, x+(1/x)= 5/2,-1/2
So, taking 5/2 as one of the root we get,
x=2,1/2. And taking -1/2 as root we get imaginary no. and which is also not given in the option.
So, the answer is x=2,1/2

Like? Yes (1) | No   

Vivek Jain   11 years AGO

let x+1/x=A
so, eqn becomes 4(A^2-2)-8*A+3
A=5/2 & -1/2
x+1/x=5/2
=>x=2,1/2
and x+1/x=-1/2 =>img roots

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Panchanand   11 years AGO

x^2+(1/x^2)={x+(1/x)}^2-2
after replacing {x^2+(1/x^2)}
equation will be
4[{x+(1/x)}^2-2]-8{x+(1/x)} + 3=0
replace {x+(1/x)} with y
now, equation will be,
4(y^2)-8y-5=0
after solving it,
y=5/2 & y=-1/2
replace value of y with {x+(1/x)} and solve for y=5/2 and we will get x=2 and x=1/2

Like? Yes (3) | No (3)  

anubhav verma   11 years AGO

(4x^3+3x-2)(x-2)=0
(2x^2+x+2)(2x-1)(x-2)=0

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Surya Narayana K   11 years AGO

Answer is 2,1/2 (Option 9).
Given that
4(x^2+(1/x^2)-8(x+1/x)+3=0
=>4x^4-8x^3+3x^2-8x+4=0
So, 2 and 1/2 are factors.

Like? Yes (1) | No (2)  

Yekollu Muneeswar   11 years AGO

4x^4-8x^3+3x^2-8x+4=0

By horner's Rule

2 |4 -8 3 -8 4
| 8 0 6 -4
--------------
4 0 3 -2 0..
2 is a solution..
similarly,

1/2| 4 -8 3 -8 4
| 2 -3 0 -4
--------------
4 -6 0 -8 0..

i.e, x=2 x=1/2 are solutions..

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mohini gupta   11 years AGO

it is a quad equation of 4 degree..
4x^4-8x^3+3x^2-8x+4=0

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Ghanshyam Gulati   11 years AGO

given eqn is
4{x^2+(1/x^2)} - 8{x+(1/x)} + 3 = 0

put x=2 in the eqn
4(2^2+(1/2^2))-8(2+(1/2))+3
on solving
we get 0
which means 2 is the root of above eqn

similarly, 1/2 also satisfy the eqn

so (2,1/2) is the ans

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ASHISH MITTAL   11 years AGO

first rearrange the equation,then equation become
4{x+(1/x)}^2 - 8{x+(1/x)} -5 = 0
let
x+(1/x)=b so equation become
4b^2-8b-5=0
solve this equation for b we get
b=2/5 and b=-1/2
after putting value of b we solve for x then we get x=2 and 1/2

Like? Yes (1) | No (2)  

DHIRENDRA    11 years AGO

4{x^2+(1/x^2)} - 8{x+(1/x)} + 3 = 0
4{X^2+1/X^2}-8{x+1/x}+3=0
4{(x^4+1)/x^2}-8{(x^2+1)/x} +3=0
4{(x^4+1)/x^2}-8{(x^2+1)*x/x^2} +3x^2/x^2=0
4(x^4+1)-8(x^3+x)+3x^2=0
4x^4+4-8x^3-8x+3x^2=0
4x^4-8x^3+3x^2-8x+4=0
putting x=2
4*2^4-8*2^3+3*2^2-8*2+4=0
4*16-64+12-16+4=0
64-64+12-16+4=0
hence x-2 is a root
put x=1/2
we will get 0
hence
root are
2,1/2

Like? Yes (1) | No (2)  

Dr.Dipin Singh   11 years AGO

if x+1/x =y, then
4{x^2+(1/x^2)} - 8{x+(1/x)} + 3 = 0
4(y^2-2)-8y+3=0
y=5/2, -1/2
so when x+1/x = 5/2
solving, we get
x=2,1/2

Like? Yes (2) | No (3)  

Rohit Agarwal   11 years AGO

we know that {x+(1/x)}^2 = x^2 + 1/x^2 + 2

so x^2 + 1/x^2=(x+(1/x))^2-2

now put this value in given equation

4{(x+(1/x))^2 - 2} - 8{x+(1/x)} + 3 =0
=> 4(x+(1/x))^2 - 8(x+(1/x))- 5=0

now let (x+(1/x))=y

4*y^2 - 8*y -5=0

by solving this equation we get y=5/2, -1/2

and we know that y=(x+(1/x))

firstly take y=5/2

so x+(1/x)=5/2

by solving this x=2,1/2

now take y=-1/2

so x+(1/x)=-1/2

this gives imaginary solutions


so our answer is x=2,1/2

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rajesh   11 years AGO

by given data solving this
get x=2,1/2

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Pallekonda Rajesh   11 years AGO

simplify the equation
we will get 4x^4-8x^3+3x^2-8x+4=0
solve this equation
ans:2,1/2

Like? Yes (3) | No (4)  

TRIMURTULU   11 years AGO

put x+(1/x)=a
then a^2={x^2+(1/x^2)}-2
then above equation 4{x^2+(1/x^2)}-8{x+(1/x)}+3= 0 becomes
4a^2-8a-5=0
if we solve it
we get 5/2 and -1/2 as solution for 'a'
i.e x+(1/x)=5/2 or x+(1/x)=-1/2
if we solve x+(1/x)=5/2
x=2 or x=1/2
Hence answer is(option 9) 2 or 1/2

Like? Yes (2) | No (3)  

VIKASH RAY   11 years AGO

4{x^2+(1/x^2)}-8{x+(1/x)}+3=0
or, 4[{x+(1/x)}^2-2]-8{x+(1/x)}+3=0
Now put {x+(1/x)}=a
then by solving the equation we get that the option 9 is correct..

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shamma mohammad   11 years AGO

by option cheking for this eqn:
4{x^2+(1/x^2)} - 8{x+(1/x)} + 3 = 0
we get 2,1/2

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basha shaik   11 years AGO

by option cheking for this eqn:
4{x^2+(1/x^2)} - 8{x+(1/x)} + 3 = 0
we get 2,1/2

Like? Yes (1) | No (2)  

naveen babu   11 years AGO

lets solve option wise
option 9)2,1/2
x=2 i.e 4(4+1/4)-8(2+1/2)+3=0
4(17/4)-8(5/2)+3=0
17-20+3=0
-3+3=0
x=1/2 i.e 4(1/4+4)-8(1/2+2)+3=0
4(17/4)-8(5/2)+3=0
17-20+3=0
-3+3=0
so option 9 is correct

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ravi jha   11 years AGO

4{x^2+(1/x^2)} - 8{x+(1/x)} + 3 = 0
by solving
x= 2,1/2

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gaurav kumar verma   11 years AGO

just by putting the values :) except this pair no one satisfies the equation

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suraj gupta   11 years AGO

In this equation,
4{x^2+(1/x^2)} - 8{x+(1/x)} + 3 = 0

Put x=2 in L.H.S.
=> 4(4+1/4)-8(2+1/2) + 3
=> 17-20+3
=> 20-20
=> 0 and equals to R.H.S.

Now, put x=1/2 in L.H.S.

=> 4(1/4+4)-8(1/2+2) + 3
=>=> 17-20+3
=> 20-20
=> 0 and equals to R.H.S.

So, x= 2,1/2

Like? Yes (1) | No (2)  

Saidinesh   11 years AGO

solving that equation
4(x^4+1)-8X(X^2+1)+3X^2=0;
4X^4-8X^3+3X^2-8X+4=0;
solving this equation
x=2,1/2..

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ch rajesh   11 years AGO

by solving get
x=2,1/2

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azharshahid01@gmail.com   11 years AGO

4{x^2+(1/x^2)} - 8{x+(1/x)} + 3 = 0
Let x+1/x=y so, (x+1/x)^2=y^2,
On solving x^2+1/x^2=y^2-2
Substitute (x+1/x) and (x^2+1/x^2) with y and y^2-2 respectively in the given equation.
Now, our equation is 4(y^2-2)-8y+3=0
On opening bracket, 4y^2-8y-5=0.
On factoring we get 4y^2-10y+2y-5=0.
On solving we get, y=5/2 and -1/2.
On substitute the value of y. So, we get (x+1)/x=5/2 and (x+1/x)=-1/2.
On solving we get, x=2 and 1/2.
So, option 9)2, 1/2 is right answer.

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