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.