x is a real number greater than 1 and y is a positive real number such that x^3 = y^2+y+1 then y

M4maths Previous Puzzles

28 Mar 2012 real and positive

x is a real number greater than 1 and y is a positive real number such that
x^3 = y^2+y+1
then y = ?

OPtion

1) [(4x^3-3)^1/2 - 1] / 2
2) [(4x^2-3)^1/2 - 1] / 2
3) [(4x^3+3)^1/2 + 1] / 2
4) [(4x^3-2)^1/2 + 1] / 2
5) [(x^3-3x)^1/2 - 1] / 2
6) [(3x^3-3)^1/2 + 1] / 2
7) [(3x^3+3)^1/3 - 1] / 2
8) [(3x^3-2)^1/2 - 1] / 2
9) [(3x^3+3)^1/2 - 1] / 2
10) none of these

Solution

x^3 = y^2+y+1
implies that x^3 - 3/4 = y^2 + y + 1/4
implies that (x^3 - 3/4)^1/2 = y + 1/2
or y = [(4x^3-3)^1/2 - 1] / 2

Correct Option: 1 Best Solution (11)

Viru Sihag   12 years AGO

y^2+y+(1-x^3)=0
then solve this quadratic equation using sridharacharya formula then we get this ans.

Like? Yes (8) | No (1)  

Balakrishnan M   12 years AGO

x^3 = y^2+y+1
multiply 4 on bothsides,
==>4x^3 = 4y^2+4y+4
==>4x^3 = 4y^2+4y+1+3
==>4x^3 =(2y+1)^2+3
==>4x^3-3 =(2y+1)^2
==>(4x^3-3)^1/2=2y+1
==>(4x^3-3)^1/2-1=2y
==>[(4x^3-3)^1/2-1]/2=y

Like? Yes (7) | No (1)  

Shiv Kumar   12 years AGO

x^3 = y^2+y+1
4x^3 = (2y)^2+4y+4 --- multiply by 4
4x^3-3 = (2y)^2+4y+1 ---- both sides minus by 3
4x^3-3 = (2y+1)^2 --- simplify RHS (a+b)^2
(4x^3-3)^1/2 = 2y+1 ---- both sides multiply in powers by 1/2

Finally [(4x^3-3)^1/2 - 1] / 2 = Y

Like? Yes (5) | No   

Rakesh Kumar Kohli   12 years AGO

x^3 = y^2+y+1
=> y^2+y+1-x^3 = 0
=> y = [-1 + (1 - 4(1-x^3))^1/2]/2 and
[-1 - (1 - 4(1-x^3))^1/2]/2
y = [-1 + (4x^3-3))^1/2]/2 and
[-1 - (4x^3-3))^1/2]/2
Since y is a positive real number so y cannot be
[-1 - (4x^3-3))^1/2]/2 which is a negative real number.
Hence y = [-1 +(4x^3-3))^1/2]/2
= [ (4x^3-3))^1/2 - 1]/2

Like? Yes (2) | No   

prithviraj sanningannavar   12 years AGO

x^3=y^2+y+1
=>y^2+y+(1-x^3)=0
This is a quadratic equation of the form ax^2+bx+c=0
Therefore the roots are
x=(-b+(sqrt(b^2-4ac)))/2a
or
x=(-b-(sqrt(b^2-4ac)))/2a

=>y=(-1+(sqrt(1-4*1*(1-x^3))))/2*1
=>y=[(4x^3-3)^1/2 - 1] / 2

So the correct answer is option 1.

Like? Yes (2) | No   

sumanth jayam   12 years AGO

By solving
y^2+y+1 = x^3 using [-b+((b^2-4*a*c)^1/2)]/2*a

where a=1,b=1,c=1-x^3 we get option 4

Like? Yes (2) | No (1)  

Prabhat Ranjan   12 years AGO

Re-arrange and solve the prob:e.g
y^2+y+(1-x^3)=0
Use the formula:y= [-b+(b^2-4ac)^1/2]/2 +ve D because of y is positive.
Therefore,
y = [-1+(1 - 4*1*(1-x^3))^1/2]/2
=> y = [(4x^3-3)^1/2 -1]/2
Hence choice 1 is correct.

Like? Yes (1) | No   

niru   12 years AGO

x3=y2+y+1
make the perfect square
x3=(y+1/2)^2 -1/4 +1
(x3-3/4)=(y+1/2)^2
[(x3-3/4)^1/2-1]/2=y
taking common 1/2 from whole we get the ans
[(4x3-3)^1/2 -1]/2=y
thats the ans

Like? Yes (2) | No (1)  

Pratik Goyal   12 years AGO

solve it by shree dharacharya method of solving quadratic equation

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Sikandar chouhan   12 years AGO

y^2+y+1-x^3=0
by solving this equation. we get [(4x^3-3)^1/2 - 1] / 2

Like? Yes (2) | No (2)  

chiru   12 years AGO

y^2+y+1-x^3=0
given that x is a real number greater than 1 and y is a positive real number hence the root must be +ve..hence the positive root of the above one is
[(4x^3-3)^1/2 - 1] / 2..therefore option 1 is the correct one..

Like? Yes (1) | No (1)