Is this right? Find real solutions to (x^2 - 2)^2 - 2 = x

[littlegentleman]

Hi. I solved this question, but i am not sure my answer is right or not.

Q. How many real solutions does the following have?

(x^2-2)^2 -2 = x

A. (x^2-2)^2 -x -2 = 0
x^4 - 4x^2 + 4 -x -2 = 0
x^4 - 4x^2 -x + 2 =0
X^2(x^2-4) -x + 2 = 0
x^2(x+2)(x-2) = (x-2) --> (x-2) cancel out from both sides
x^2(x+2) = 1
x^2 = 1 or (x+2) = 1
Therefore, the answer is "it has two solutions, x = -1, 1."

Please tell me if i did something wrong.
Thank you very much.

 

[Deleted member 4993]

Re: Is this right?

Hi. I solved this question, but i am not sure its right or not.
Q. How many real solutions does the following have?
(x^2-2)^2 -2 = x
A. (x^2-2)^2 -x -2 = 0
x^4 - 4x^2 + 4 -x -2 = 0
x^4 - 4x^2 -x + 2 =0
X^2(x^2-4) -x + 2 = 0
x^2(x+2)(x-2) = (x-2) --> (x-2) cancel out from both sides - should not do that - you are loosing one solution there x^2(x+2) = 1
x^2 = 1 or (x+2) = 1<--- This step is not correct.
Now you have

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

from

x-2 = 0

One solution is x = 2

Other solutions will come from the "cubic" factor.
Therefore, the answer is "it has two solutions, x = -1, 1."
Please tell me if i did something wrong.
Thank you very much.

One way to check your solution is to plug those back into the equations and see if those satisfy the equation.

Actually - it has 4 solutions.

What level of math are doing?

If you are in high school - I'll advise to solve it graphically.

Plot y = (x^2-2)^2 and y = x+2

Find the points of intersections of these two graphs.

 

[littlegentleman]

Re: Is this right?

Hi. Even though I graph it, I can only find 2 answers. x = 2 and -1.
And if I solve (x^3)+(2x^2)-1=0
(x+1)(x^2+x-1) = 0
From (x^2+x-1) = 0
x = -b +-[squareroot b^2-4ac] / 2a
Therefore, x = -1 + [squareroot 5] / 2 & -1 - [squareroot 5] / 2
Is this correct? Please tell me if i did something wrong.
Thank you very much.

 

[Deleted member 4993]

Re: Is this right?

Hi. Even though I graph it, I can only find 2 answers. x = 2 and -1.
And if I solve (x^3)+(2x^2)-1=0
(x+1)(x^2+x-1) = 0
From (x^2+x-1) = 0
x = -b +-[squareroot b^2-4ac] / 2a
Therefore, x = -1 + [squareroot 5] / 2 & -1 - [squareroot 5] / 2

Check your answer by putting these values into the original equation and see if the equation is satisfied.

Is this correct? Please tell me if i did something wrong.
Thank you very much.