Question

The number of real roots of the equation $ x|x-2| + 3|x-3| + 1 = 0 $ is:

• A. 4
• B. 2
• C. 1
• D. 3

Hint:

When solving absolute value equations, consider different cases based on the sign of the expression inside the absolute value. This will help simplify the equation and allow for easier solving.

Answer 1

The Correct Option is C

We are given the equation: \[ x|x-2| + 3|x-3| + 1 = 0 \]
Step 1: Analyze the absolute value functions
The absolute value expressions depend on the values of \(x\). Let’s consider the cases based on \(x\).
Case 1: \(x \geq 3\)
In this case:
\( |x-2| = x - 2 \) \( |x-3| = x - 3 \) The equation becomes: \[ x(x-2) + 3(x-3) + 1 = 0 \] Simplifying: \[ x^2 - 2x + 3x - 9 + 1 = 0 \] \[ x^2 + x - 8 = 0 \] Solving the quadratic equation using the discriminant: \[ x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-8)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 32}}{2} = \frac{-1 \pm \sqrt{33}}{2} \] This gives two real solutions, but only one of them is in the range \(x \geq 3\). Therefore, this case gives
1 solution
.
Case 2: \(2 \leq x<3\)
Here, we have: \( |x-2| = x - 2 \) \( |x-3| = 3 - x \) Substitute into the equation: \[ x(x-2) + 3(3-x) + 1 = 0 \] Simplifying: \[ x^2 - 2x + 9 - 3x + 1 = 0 \] \[ x^2 - 5x + 10 = 0 \] The discriminant for this quadratic is: \[ \Delta = (-5)^2 - 4(1)(10) = 25 - 40 = -15 \] Since the discriminant is negative, there are
no real solutions
in this case.
Case 3: \(x<2\)
In this case: \( |x-2| = 2 - x \) \( |x-3| = 3 - x \) Substitute into the equation: \[ x(2 - x) + 3(3 - x) + 1 = 0 \] Simplifying: \[ 2x - x^2 + 9 - 3x + 1 = 0 \] \[ -x^2 - x + 10 = 0 \] Solving: \[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(-1)(10)}}{2(-1)} = \frac{1 \pm \sqrt{1 + 40}}{-2} = \frac{1 \pm \sqrt{41}}{-2} \] The discriminant is positive, so there are
2 real solutions
in this case.
Final Answer:
The total number of real solutions is \(1 \text{ solution from Case 1}\). Thus, the total number of real roots of the equation is \(1\).