Find all real solutions to the equation?

Question

x^2 + x + sqrt[x^2 + x] - 2 = 0


I am just plain confused about eliminating the square root, etc.

Answer 1

Let u = sqrt(x^2 + x). Your equation becomes u^2 + u - 2 = 0. Have fun.

Answer 2

x^2 + x + sqrt[x^2 + x] - 2 = 0
or x(x+1)+ sqrt[x(x+1)] - 2 = 0 (1)
put sqrt[ x(x+1)] = y in (1)
therefore (1) becomes
or y^2+ y - 2 = 0 (2)
or (y+2)(y-1) =0 (3)
implies y =-2 or y =1
Now
when y = - 2 ,
but sqrt x(x+1) = y
implies sqrt [x(x+1) ]= -2
x(x+1) = 4 on squaring
implies x^2+x-4 = 0 (4)
Solving (4)
x = { -1+sqrt [1-4*1*(-4)]}/2 = { -1+sqrt [17]}/2 (5)
x = { -1+sqrt [1-4*1*(-4)]}/2 = { -1-sqrt [117]}/2 (6)

Similarly
when y = 1,
but sqrt x(x+1) =1
implies sqrt [x(x+1) ]= 1
x(x+1) = 1 on squaring
implies x^2+x-1 = 0 (7)
Solving (7)
x = { -1+sqrt [1-4*1*(-1)]}/2 = { -1+sqrt [5]}/2 (8)
x = { -1+sqrt [1-4*1*(-1)]}/2 = { -1-sqrt [5]}/2 (9)
Hence (5),(6) ,(8) and (9) are the real solutions of the equation

Answer 3

Hi Liberty,

The trick is to substitute.

Let y=x^2 + x

now we have y + Sqrt (y) = 0

separate the two.

y=sqrt y

Square both sides

y^2 = y

So

y^2 - y = 0

y(y-1) = 0

So either y = 0, or y=1

Now substituting back,

x^2 + x = 0

or

x^2 + x = 1


For
x^2 + x = 0
x(x+1) = 0

x = 0, or -1

For
x^2 + x = 1
x^2 + x -1 = 0

using the quadratic equation,

x = (-1 +- Sqrt (1 + 4))/2

x = (-1 +- Sqrt5)/2

There are your four answers. Hope that helps,
Matt

Answer 4

im pretty sure you get rid of a square root by multiplying both sides by 2

Answer 5

sqrt(x^2+x) = 2-x-x^2
x^2+x = 4-4x-3x^2+2x^3+x^4
x^4 +2x^3 -4x^2-5x +4 = 0
You can take it from here.