What can you say about the equation (x^2 + 1) = 2x - x^2 - 2 ?

Question

a) it has x=2 as a solution.
b) it has no real solutions.
c) it has an odd number of real solutions.
d) it has 20 real solutions.

Please explain how you did it.

Sorry I meant to say (x^2 + 1)^10

Answer 1

You can use the same method for (x^2 + 1)^10
Well I would start by saying (x^2 + 1)^10 ≥ 1.
This is because:
x^2 ≥ 0
Therefore x^2+1 ≥ 1
So, (x^2 + 1)^10 ≥ 1.


Since it is an equation, this neccessarily means that:
2x - x^2 -2 ≥ 1 (because if the LHS ≥ 1; the RHS ≥ 1)
Rearrange this to get:
x^2 - 2x + 3 ≤ 0
Complete the square (this is just writing the LHS in another form that happens to be very helpful in a lot of cases):
(x-1)^2 + 2 ≤ 0
So (x-1)^2 ≤ -2
Therefore there are no real solutions, because the square root of -2 is imaginary.

Thus the answer is (b).
ps LHS means 'left hand side' and RHS means, well guess for yourself.

Answer 2

Combine like variables
2x^2 - 2x + 3 =0
Use the quadratic formula 'cuz I can't factor
it and neither can anyone else

x= 2(+/-)√(4-24)/4
x=2(+/-)√(-20)/4 and as you can see there's a
-20 in the radical so you get imaginary roots.
The answer is b), no real roots.

Answer 3

x^2+1-2x+x^2+2=0
2x^2-2x+3=0
D = b^2-4ac = (-2)^2-4.2.3=4-24 = -20

D<0 means that this equation has no real sol.
(if D>0, it means it has 2 real different sol, and if D=0 it means it has 2 same real sol)

so the answer is b