Question

(x^2 +x-6)^2+|x+3|=0

Answer 1

(x²+x-6)²≥0

|x+3|≥0

Cумма неотрицательных слагаемых равна 0 тогда и только тогда, когда каждое 0

$\left \{ {{x^2+x-6=0} \atop {x+3=0}} \right. \left \{ {{D=25; x_{1}=-3;x_{2}=2} \atop {x=-3}} \right.$

О т в е т. х=-3

Answer 2

Let's solve the equation step by step:

1. Start by simplifying the expression: (x^2 + x - 6)^2 + |x + 3| = 0

2. Since the expression (x^2 + x - 6)^2 is always non-negative (its square is always positive or zero), we can ignore it for now and focus on the absolute value term: |x + 3| = 0

3. The absolute value of a number is zero if and only if the number itself is zero. Therefore, we have x + 3 = 0.

4. Solve the equation x + 3 = 0 for x: Subtract 3 from both sides: x = -3

5. Now, let's consider the remaining expression (x^2 + x - 6)^2 and see if it can be equal to zero.

6. (x^2 + x - 6)^2 is a square, and squares are always non-negative. So, for the expression to be zero, the squared term must be zero: x^2 + x - 6 = 0.

7. Solve the quadratic equation x^2 + x - 6 = 0. We can factor it as (x + 3)(x - 2) = 0.

8. Set each factor equal to zero and solve for x: x + 3 = 0 --> x = -3 x - 2 = 0 --> x = 2

9. Therefore, we have three solutions: x = -3, x = -3, and x = 2.

Please note that one of the solutions, x = -3, was obtained from the absolute value term |x + 3| = 0. However, since absolute value is always non-negative, the equation |x + 3| = 0 only has one solution, which is x = -3.