Question

|x+1|+|x-2|=3(x-1)^2-2|x-1|=3​

Answer 1

1) х€[-1;2]
2) х1=-2. х2=4

Answer 2

Let's solve the given equation step by step:

|x + 1| + |x - 2| = 3(x - 1)^2 - 2|x - 1| = 3

To simplify the equation, let's consider two cases:

Case 1: x < 1 In this case, both (x + 1) and (x - 2) are negative. So we can rewrite the equation as follows:

-(x + 1) - (x - 2) = 3(x - 1)^2 - 2(x - 1)

• x - 1 - x + 2 = 3(x - 1)^2 - 2x + 2
• 2x + 1 = 3(x - 1)^2 - 2x + 2
• 2x + 1 = 3x^2 - 6x + 3 - 2x + 2
• 2x + 1 = 3x^2 - 8x + 5

Move all the terms to one side:

3x^2 - 6x + 3x - 8x + 2x + 1 - 5 = 0 3x^2 - 9x - 4 = 0

Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a), where a = 3, b = -9, and c = -4:

x = (-(-9) ± √((-9)^2 - 4 * 3 * (-4))) / (2 * 3) x = (9 ± √(81 + 48)) / 6 x = (9 ± √129) / 6

So for x < 1, the solutions are: x = (9 + √129) / 6 x = (9 - √129) / 6

Case 2: x ≥ 1 In this case, both (x + 1) and (x - 2) are positive. So we can rewrite the equation as follows:

(x + 1) + (x - 2) = 3(x - 1)^2 - 2(x - 1) x + 1 + x - 2 = 3(x - 1)^2 - 2x + 2 2x - 1 = 3x^2 - 6x + 3 - 2x + 2 2x - 1 = 3x^2 - 8x + 5

Move all the terms to one side:

3x^2 - 6x + 3x - 8x + 2x - 1 - 5 = 0 3x^2 - 9x - 3 = 0

Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a), where a = 3, b = -9, and c = -3:

x = (-(-9) ± √((-9)^2 - 4 * 3 * (-3))) / (2 * 3) x = (9 ± √(81 + 36)) / 6 x = (9 ± √117) / 6

So for x ≥ 1, the solutions are: x = (9 + √117) / 6 x = (9 - √117) / 6

Therefore, the complete set of solutions to the equation |x + 1| + |x - 2| = 3(x - 1)^2 - 2|x - 1| = 3 is:

x = (9 + √129) / 6 x = (9 - √129) / 6 x = (9 + √117) / 6 x = (9 - √117) / 6