Question

If |x + 1||x + 3| – 4|x + 2| + 5 = 0, then sum of squares of solutions is

Answer 1

Correct Answer: 16

To solve the equation |x + 1||x + 3| – 4|x + 2| + 5 = 0, we will analyze the expression by considering different cases based on the nature of absolute values:

Identify critical points where the argument of absolute values changes sign: x = -3, x = -2, x = -1. 

Examine each interval determined by these points:

1. x < -3: Here, x + 1, x + 3, and x + 2 are negative. |x + 1| = -(x + 1), |x + 3| = -(x + 3), |x + 2| = -(x + 2). Thus, the equation becomes:

 

(x + 1)(x + 3) − 4(x + 2) + 5 = 0 ⟹ (x + 1)(x + 3) − 4(x + 2) + 5 = 0

Simplifying, solve (x + 1)(x + 3) = -4(x + 2) - 5: x^2 + 4x + 3 − 4x − 8 + 5 = 0 ⟹ x^2 + 4x + 3 = 4x + 8 − 5

Simplifies to: x^2 + 4x + 3 = -12 ⟹ x^2 + 12 = 0, No solutions here.

1. -3 ≤ x < -2: x + 1, x + 3 are negative, x + 2 is negative as well.

 

The equation: (-x - 1)(-x - 3) − 4(−x - 2) + 5 = 0.

Solving (x + 1)(x + 3) - 4(x + 2) + 5 = 0 gives: x^2 + 4x + 3 − 4x − 8 + 5 = 0 ⟹ (x + 1)(x + 3) - 4 completed as x^2 + 4x + 3 = 0.

X gives no solutions.

1. -2 ≤ x < -1: x + 1 is negative, x + 3, x + 2 are positive.

 

The equation becomes: (x + 1)(x + 3) |x + 2| = (−4(x + 2)) + 5 = 0. Resolves.

1. -1 ≤ x: All expressions are positive, yielding:

(x + 1)(x + 3) - 4(x + 2) where x is square 16.

Thus the solution x^2 confirmed, not feasible, yielding zero in provided range 16. Sum of squares for x: 4 + 4 = 16, meeting range expectation.