The given equation is x(x^2 + 3|x| + 5|x-1| + 6|x-2|) = 0.

This equation will have real solutions when either x = 0 or when the expression in the parentheses equals zero.

When x = 0, the equation is satisfied, so x = 0 is one real solution.

For the expression in the parentheses to equal zero, we need to consider the absolute value expressions separately for positive and negative values of x.

a. When x > 0, the equation becomes x(x^2 + 3x + 5(x-1) + 6(x-2)) = 0. Simplifying this gives x(x^2 + 3x + 5x - 5 + 6x - 12) = 0, which simplifies further to x(x^2 + 14x - 17) = 0. This quadratic equation has two real solutions, but since we assumed x > 0, only the positive solution is valid.

b. When x < 0, the equation becomes x(x^2 - 3x - 5(x-1) - 6(x-2)) = 0. Simplifying this gives x(x^2 - 3x - 5x + 5 - 6x + 12) = 0, which simplifies further to x(x^2 - 14x + 17) = 0. This quadratic equation also has two real solutions, but since we assumed x < 0, only the negative solution is valid.

So, in total, the equation has 1 (from x = 0) + 1 (from x > 0) + 1 (from x < 0) = 3 real solutions.

Question

The given equation is x(x^2 + 3|x| + 5|x-1| + 6|x-2|) = 0.

This equation will have real solutions when either x = 0 or when the expression in the parentheses equals zero.

When x = 0, the equation is satisfied, so x = 0 is one real solution.
For the expression in the parentheses to equal zero, we need to consider the absolute value expressions separately for positive and negative values of x.

a. When x > 0, the equation becomes x(x^2 + 3x + 5(x-1) + 6(x-2)) = 0. Simplifying this gives x(x^2 + 3x + 5x - 5 + 6x - 12) = 0, which simplifies further to x(x^2 + 14x - 17) = 0. This quadratic equation has two real solutions, but since we assumed x > 0, only the positive solution is valid.

b. When x < 0, the equation becomes x(x^2 - 3x - 5(x-1) - 6(x-2)) = 0. Simplifying this gives x(x^2 - 3x - 5x + 5 - 6x + 12) = 0, which simplifies further to x(x^2 - 14x + 17) = 0. This quadratic equation also has two real solutions, but since we assumed x < 0, only the negative solution is valid.

So, in total, the equation has 1 (from x = 0) + 1 (from x > 0) + 1 (from x < 0) = 3 real solutions.

Knowee AI · Accepted Answer

The given equation is x(x^2 + 3|x| + 5|x-1| + 6|x-2|) = 0.

This equation will have real solutions when either x = 0 or when the expression in the parentheses equals zero.

When x = 0, the equation is satisfied, so x = 0 is one real solution.
For the expression in the parentheses to equal zero, we need to consider the absolute value expressions separately for positive and negative values of x.

a. When x > 0, the equation becomes x(x^2 + 3x + 5(x-1) + 6(x-2)) = 0. Simplifying this gives x(x^2 + 3x + 5x - 5 + 6x - 12) = 0, which simplifies further to x(x^2 + 14x - 17) = 0. This quadratic equation has two real solutions, but since we assumed x > 0, only the positive solution is valid.

b. When x < 0, the equation becomes x(x^2 - 3x - 5(x-1) - 6(x-2)) = 0. Simplifying this gives x(x^2 - 3x - 5x + 5 - 6x + 12) = 0, which simplifies further to x(x^2 - 14x + 17) = 0. This quadratic equation also has two real solutions, but since we assumed x < 0, only the negative solution is valid.

So, in total, the equation has 1 (from x = 0) + 1 (from x > 0) + 1 (from x < 0) = 3 real solutions.

The number of real solutions of the equation x(x2+3|x|+5|x−1|+6|x−2|)=0 is_______.