Question

Решите |х-х^2-1|=|2х-3+х^2|

Answer 1

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Готово.
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Answer 2

To solve the equation |x - x^2 - 1| = |2x - 3 + x^2|, we need to consider different cases based on the sign of the expressions inside the absolute value bars.

Case 1: (x - x^2 - 1) and (2x - 3 + x^2) are both non-negative.

In this case, the equation simplifies to: x - x^2 - 1 = 2x - 3 + x^2

Now, let's solve for x: x^2 + x^2 - x - 2x + 3 - 1 = 0 2x^2 - 3x + 2 = 0

Using the quadratic formula: x = (-(-3) ± √((-3)^2 - 4 * 2 * 2)) / (2 * 2) x = (3 ± √(9 - 16)) / 4 x = (3 ± √(-7)) / 4

Since the discriminant (9 - 16) is negative, the roots are complex, and there are no real solutions in this case.

Case 2: (x - x^2 - 1) is non-negative, and (2x - 3 + x^2) is negative.

In this case, the equation simplifies to: x - x^2 - 1 = -(2x - 3 + x^2)

Now, let's solve for x: x - x^2 - 1 = -2x + 3 - x^2

Rearrange the equation: x^2 - x - 4 = 0

Using the quadratic formula: x = (1 ± √(1^2 - 4 * 1 * (-4))) / 2 x = (1 ± √(1 + 16)) / 2 x = (1 ± √17) / 2

So, in this case, we have two solutions:

1. x = (1 + √17) / 2 ≈ 1.56155
2. x = (1 - √17) / 2 ≈ -0.56155

Case 3: (x - x^2 - 1) is negative, and (2x - 3 + x^2) is non-negative.

In this case, the equation simplifies to: -(x - x^2 - 1) = 2x - 3 + x^2

Now, let's solve for x: x^2 - x + 4 = 0

Using the quadratic formula: x = (1 ± √(1^2 - 4 * 1 * 4)) / 2 x = (1 ± √(1 - 16)) / 2 x = (1 ± √(-15)) / 2

Since the discriminant (-15) is negative, the roots are complex, and there are no real solutions in this case.

In conclusion, the real solutions to the equation |x - x^2 - 1| = |2x - 3 + x^2| are approximately:

1. x ≈ 1.56155
2. x ≈ -0.56155