How many (real) solutions do the equation #abs(x-1) = x^2 + 1# have?

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Answer 1 · 2018-05-17T06:21:35

3. #2#

Given:

#abs(x-1) = x^2+1#

Trying a few small values, we can find solutions:

#abs((color(blue)(0))-1) = 1 = (color(blue)(0))^2+1#

#abs((color(blue)(-1))-1) = 2 = (color(blue)(-1))^2+1#

So it looks like there are #2# solutions.

More formally and to check we have not missed any, let's try squaring the given equation to get (for real values of #x#):

#x^2-2x+1 = x^4+2x^2+1#

Then subtracting #x^2-2x+1# from both sides we get:

#0 = x^4+x^2+2x#

#color(white)(0) = x(x^3+x+2)#

#color(white)(0) = x(x+1)(x^2-x+2)#

The remaining quadratic is always positive as we can see by checking its discriminant or by completing the square:

#x^2-x+2 = (x-1/2)^2+7/4#

So it has no factors with real coefficients.