The number of real solution(s) of the equation x^2 + 3x + 2 = min{|x - 3|, |x + 2|} is:
(1) 1
(2) 0

Q: Why do we split the function at $x = 0.5$? Can't we just solve the equation directly?

The $\min$ function is a piecewise function. It behaves differently depending on which of the two expressions, $|x-3|$ or $|x+2|$, is smaller. By finding the intersection point $x = 0.5$, we identify the exact boundary where the 'lower path' switches from one function to the other. Without this split, you cannot algebraically remove the $\min$ operator.

Q: Why did we reject the roots in Case 2a? They are valid solutions to the quadratic equation, right?

They are valid solutions to the *quadratic equation* we derived, but they are not valid solutions to the *original equation* within that specific interval. In Case 2a, we assumed $x \in [0.5, 3)$. If a solution falls outside this range, it means our assumption for that case was incorrect, and that value of $x$ does not satisfy the condition required for the $\min$ function to behave as $|x-3|$ in that region.

Q: Is there a faster way to solve this, like squaring both sides?

Squaring both sides of an equation involving absolute values is a dangerous trap. It often introduces extraneous roots that don't satisfy the original equation. While it might seem faster, the piecewise approach is the most rigorous and reliable method for JEE Advanced problems involving absolute values and minimum/maximum functions.

Question

The number of real solution(s) of the equation $x^{2}+3x+2=\min\{|x-3|,|x+2|\}$ is:

Options:

(A) 1

(B) 0

(C) 2

(D) 3

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with finding the number of real solutions to the equation:

$$x^2 + 3x + 2 = \min\{|x-3|, |x+2|\}$$

Let $f(x) = x^2 + 3x + 2$. Factoring this quadratic, we get $f(x) = (x+1)(x+2)$, which represents an upward-opening parabola with roots at $x = -1$ and $x = -2$.

Let $g(x) = \min\{|x-3|, |x+2|\}$. This function represents the lower envelope of two V-shaped absolute value functions, one with a vertex at $x = -2$ and the other at $x = 3$.

### Finding the Switching Point

To define $g(x)$ explicitly, we identify where the two absolute value functions intersect:

$$|x-3| = |x+2|$$

Setting $-(x-3) = x+2$ (since the intersection occurs between the vertices), we obtain $3 - x = x + 2$, which simplifies to $2x = 1$, or $x = 0.5$.

Thus, the function $g(x)$ is defined piecewise as:

$$g(x) = 
\begin{cases} 
|x+2| & 	ext{if } x < 0.5 \
|x-3| & 	ext{if } x \ge 0.5 
\end{cases}$$

### Case-by-Case Analysis

We now solve the equation $f(x) = g(x)$ by examining the defined intervals.

**Case 1: $x < 0.5$**

In this region, the equation is $x^2 + 3x + 2 = |x+2|$. We further split this based on the sign of $x+2$:

1. **Sub-case 1a ($x \in [-2, 0.5)$):** Here, $|x+2| = x+2$. The equation becomes $x^2 + 3x + 2 = x + 2$, which simplifies to $x^2 + 2x = 0$. Factoring gives $x(x+2) = 0$, yielding **$x = 0$** and **$x = -2$**. Both are valid solutions.

2. **Sub-case 1b ($x < -2$):** Here, $|x+2| = -(x+2)$. The equation becomes $x^2 + 3x + 2 = -x - 2$, leading to $x^2 + 4x + 4 = 0$, or $(x+2)^2 = 0$. This gives $x = -2$, which is not in the strict interval $x < -2$.

**Case 2: $x \ge 0.5$**

In this region, the equation is $x^2 + 3x + 2 = |x-3|$. We split this based on the sign of $x-3$:

1. **Sub-case 2a ($x \in [0.5, 3)$):** Here, $|x-3| = 3-x$. The equation becomes $x^2 + 3x + 2 = 3 - x$, or $x^2 + 4x - 1 = 0$. Using the quadratic formula:

$$x = \frac{-4 \pm \sqrt{16 - 4(1)(-1)}}{2} = -2 \pm \sqrt{5}$$

Since $\sqrt{5} \approx 2.236$, the roots are $\approx 0.236$ and $\approx -4.236$. Neither value lies in the interval $[0.5, 3)$.

2. **Sub-case 2b ($x \ge 3$):** Here, $|x-3| = x-3$. The equation becomes $x^2 + 3x + 2 = x - 3$, or $x^2 + 2x + 5 = 0$. The discriminant is $\Delta = 2^2 - 4(1)(5) = -16$. Since $\Delta < 0$, there are no real roots in this interval.

### Final Conclusion

After evaluating all possible intervals, we find that the only valid solutions are $x = 0$ and $x = -2$.

The total number of real solutions is **2**.

Animated Solution for Mathematics - Quadratic Equations: The number of real solution(s) of the equation $x^{2} + 3 x + 2 = min {∣ x - 3∣, ∣ x + 2∣}$ is:

Select Answer:

Visualized Solution

Defining the Functions

Analyzing $f (x)$

Analyzing the Components of $g (x)$

Finding the Intersection of Absolute Functions

Defining $g (x)$ Piecewise

Case 1: Solving for $x < 0.5$

Case 1a: $x \in [- 2, 0.5)$

Case 1b: $x < - 2$

Case 2: Solving for $x \geq 0.5$

Case 2a: $x \in [0.5, 3)$

Case 2b: $x \geq 3$

Final Conclusion

The Sigma Insight: Solution of Quadratic Equations

Analyzing the Setup

Finding the Switching Point

Case-by-Case Analysis

Final Conclusion

Similar Questions

The number of real solutions of the equation $∣ x ∣^{2} - 3∣ x ∣ + 2 = 0$ is

The number of real solutions of the equation $3 (x^{2} + \frac{1}{x ^{2}}) - 2 (x + \frac{1}{x}) + 5 = 0$ , is

The number of real solutions of the equation $x ∣ x + 5∣ + 2∣ x + 7∣ - 2 = 0$ is_________

The number of distinct real solutions of the equation $x ∣ x + 4∣ + 3∣ x + 2∣ + 10 = 0$ is

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For $x \in R$ , then number of real roots of the equation $3 x^{2} - 4∣ x^{2} - 1∣ + x - 1 = 0$ is ____.

The number of distinct real roots of the equation $∣ x + 1∣∣ x + 3∣ - 4∣ x + 2∣ + 5 = 0$ , is

The number of the real roots of the equation $(x + 1)^{2} + ∣ x - 5∣ = \frac{27}{4}$ is

Animated Solution for Mathematics - Quadratic Equations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of real solution(s) of the equation x2+3x+2=min{∣x−3∣,∣x+2∣} is:

Select Answer:

Visualized Solution

Defining the Functions

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Analyzing the Components of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

Finding the Intersection of Absolute Functions

Defining .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x) Piecewise

Case 1: Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x<0.5

Case 1a: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x∈[−2,0.5)

Case 1b: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x<−2

Case 2: Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x≥0.5

Case 2a: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x∈[0.5,3)

Case 2b: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x≥3

Final Conclusion

The Sigma Insight: Solution of Quadratic Equations

Analyzing the Setup

Finding the Switching Point

Case-by-Case Analysis

Final Conclusion

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Animated Solution for Mathematics - Quadratic Equations: The number of real solution(s) of the equation $x^{2} + 3 x + 2 = min {∣ x - 3∣, ∣ x + 2∣}$ is:

Analyzing $f (x)$

Analyzing the Components of $g (x)$

Defining $g (x)$ Piecewise

Case 1: Solving for $x < 0.5$

Case 1a: $x \in [- 2, 0.5)$

Case 1b: $x < - 2$

Case 2: Solving for $x \geq 0.5$

Case 2a: $x \in [0.5, 3)$

Case 2b: $x \geq 3$

The number of real solutions of the equation $∣ x ∣^{2} - 3∣ x ∣ + 2 = 0$ is

The number of real solutions of the equation $3 (x^{2} + \frac{1}{x ^{2}}) - 2 (x + \frac{1}{x}) + 5 = 0$ , is

The number of real solutions of the equation $x ∣ x + 5∣ + 2∣ x + 7∣ - 2 = 0$ is_________

The number of distinct real solutions of the equation $x ∣ x + 4∣ + 3∣ x + 2∣ + 10 = 0$ is

The number of real solutions of the equation, $x^{2} - ∣ x ∣ - 12 = 0$ is:

The number of real roots of the equation $x ∣ x ∣ - 5∣ x + 2∣ + 6 = 0$ , is

The number of real roots of the equation $x^{2} - 4 x + 3 + x^{2} - 9 = 4 x^{2} - 14 x + 6$ , is:

For $x \in R$ , then number of real roots of the equation $3 x^{2} - 4∣ x^{2} - 1∣ + x - 1 = 0$ is ____.

The number of distinct real roots of the equation $∣ x + 1∣∣ x + 3∣ - 4∣ x + 2∣ + 5 = 0$ , is

The number of the real roots of the equation $(x + 1)^{2} + ∣ x - 5∣ = \frac{27}{4}$ is