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

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Answer 1

Eva Adamson

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.

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Answer 2

Savannah Anderson

The equation ( |x-1| = x^2 + 1 ) has 2 real solutions.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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

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