Question

If the equation \[ |x^2 - 6x + 8| = a \] \(\text{has four real solutions, then find the value of \( a \):}\)
 

• A. \( a \in [0, \infty) \)
• B. \( a = 1 \)
• C. \( a \in (0,1) \)
• D. \( a \in [1,2] \)

Hint:

When dealing with absolute value equations, always split them into two cases and ensure the discriminant is positive for real solutions to exist.

Answer 1

The Correct Option is C

We are given the equation: \[ |x^2 - 6x + 8| = a \] This absolute value equation will split into two cases: Case 1: \( x^2 - 6x + 8 = a \) This is a quadratic equation: \[ x^2 - 6x + (8 - a) = 0 \] The discriminant (\( \Delta \)) of this quadratic equation must be non-negative for real solutions to exist: \[ \Delta = (-6)^2 - 4(1)(8 - a) = 36 - 4(8 - a) = 36 - 32 + 4a = 4 + 4a \] For real solutions, the discriminant must be greater than or equal to zero: \[ 4 + 4a \geq 0 \] \[ a \geq -1 \] This is a necessary condition, but to ensure four real solutions, we need the equation to have two distinct real roots, which means \( \Delta > 0 \), i.e., \[ a > 0 \] Case 2: \( x^2 - 6x + 8 = -a \) This is another quadratic equation: \[ x^2 - 6x + (8 + a) = 0 \] The discriminant for this case is: \[ \Delta = (-6)^2 - 4(1)(8 + a) = 36 - 4(8 + a) = 36 - 32 - 4a = 4 - 4a \] For real solutions, we need the discriminant to be non-negative: \[ 4 - 4a \geq 0 \] \[ a \leq 1 \] Conclusion: For the equation to have four real solutions, we need both conditions: 1. \( a > 0 \) (from Case 1), 2. \( a \leq 1 \) (from Case 2). Thus, the value of \( a \) must lie in the range: \[ 0 < a \leq 1 \] Therefore, the correct answer is \( \boxed{(c)} \).