The number of real roots of the equation
$
x|x-2| + 3|x-3| + 1 = 0
$
is:
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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.
To determine the number of real roots for the equation:
\(x|x-2| + 3|x-3| + 1 = 0\)
we need to analyze the function through its critical points. Specifically, we observe where each absolute value expression changes from positive to negative or vice versa. This will occur at the points where the expressions inside the absolute values are zero. The critical points are:
\(x = 2\) (for \(|x-2|\))
\(x = 3\) (for \(|x-3|\))
We will evaluate the behavior of the equation over the intervals formed by these critical points: \(( -\infty, 2 )\), \(( 2, 3 )\), and \(( 3, \infty )\).
**Interval** \((-\infty, 2)\): For \(x < 2\), both \(|x-2| = 2-x\) and \(|x-3| = 3-x\). The equation becomes: \(x(2-x) + 3(3-x) + 1 = 0\) simplifies to: \(-x^2 + 5x - 8 = 0\). The discriminant, \(b^2 - 4ac\), is negative: \(25 - 32 = -7\), indicating no real roots in this interval.
**Interval** \((2, 3)\): In this region, \(|x-2| = x-2\) and \(|x-3| = 3-x\). The equation becomes: \(x(x-2) + 3(3-x) + 1 = 0\), simplifying to: \(x^2 - 5x + 10 = 0\). The discriminant is \(25 - 40 = -15\). Hence, no real roots occur here.
**Interval** \((3, \infty)\): Here, \(|x-2| = x-2\) and \(|x-3| = x-3\). The equation becomes: \(x(x-2) + 3(x-3) + 1 = 0\), simplifying to: \(x^2 + x - 10 = 0\). The discriminant is positive: \(1 + 40 = 41\), which means two distinct roots exist.
The real roots in \((3, \infty)\) need to be verified for being actual, keeping in mind that only one of them might be greater than 3. By evaluating, we find exactly one real root in this range.
Hence, the number of real roots of the equation is 1.
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Approach Solution -2
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 \(1\).
