Question

The number of real solution of equation x$|$x+4$|$+3$|$x+2$|$+10=0 is/are :

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

When solving equations with multiple absolute value terms, always identify the critical points (where the arguments become zero) and analyze the equation in the intervals defined by these points.

Answer 1

The Correct Option is A

Step 1: Understanding the Equation
The given equation involves absolute value functions: \[ x|x+4| + 3|x+2| + 10 = 0 \] To solve this, we will break the equation down into separate cases based on the values of \(x\) that make the expressions inside the absolute values change signs. We need to analyze this in different intervals defined by the critical points where the arguments inside the absolute values are zero. 

Step 2: Identify Critical Points and Define Intervals
The two absolute value functions are \( |x+4| \) and \( |x+2| \). The arguments inside the absolute values become zero at: \[ x+4 = 0 \quad \Rightarrow \quad x = -4 \] and \[ x+2 = 0 \quad \Rightarrow \quad x = -2 \] These points divide the number line into three intervals: 
Case 1: \( x<-4 \) 
Case 2: \( -4 \leq x<-2 \) 
Case 3: \( x \geq -2 \) 
Step 3: Analyze and Solve the Equation in Each Interval
Case 1: \( x<-4 \)
In this interval, both \( x+4<0 \) and \( x+2<0 \). Therefore, the absolute values become: \[ |x+4| = -(x+4) \quad \text{and} \quad |x+2| = -(x+2) \] Substituting these into the equation: \[ x(-(x+4)) + 3(-(x+2)) + 10 = 0 \] Simplifying: \[ -x(x+4) - 3(x+2) + 10 = 0 \] \[ -x^2 - 4x - 3x - 6 + 10 = 0 \] \[ -x^2 - 7x + 4 = 0 \] Rewriting this as: \[ x^2 + 7x - 4 = 0 \] We can solve this quadratic equation using the quadratic formula: \[ x = \frac{-7 \pm \sqrt{7^2 - 4(1)(-4)}}{2(1)} = \frac{-7 \pm \sqrt{49 + 16}}{2} = \frac{-7 \pm \sqrt{65}}{2} \] The two possible solutions are: \[ x_1 = \frac{-7 + \sqrt{65}}{2}, \quad x_2 = \frac{-7 - \sqrt{65}}{2} \] Approximating \(\sqrt{65} \approx 8.06\), we get: \[ x_1 \approx \frac{-7 + 8.06}{2} = \frac{1.06}{2} = 0.53 \quad (\text{Not in the interval} \, x<-4) \] \[ x_2 \approx \frac{-7 - 8.06}{2} = \frac{-15.06}{2} = -7.53 \quad (\text{In the interval} \, x<-4) \] Thus, the solution from this case is: \[ x = \frac{-7 - \sqrt{65}}{2} \] 
Case 2: \( -4 \leq x<-2 \)
In this interval, \( x+4 \geq 0 \) and \( x+2<0 \), so the absolute values become: \[ |x+4| = x+4 \quad \text{and} \quad |x+2| = -(x+2) \] Substituting these into the equation: \[ x(x+4) + 3(-(x+2)) + 10 = 0 \] Simplifying: \[ x^2 + 4x - 3x - 6 + 10 = 0 \] \[ x^2 + x + 4 = 0 \] The discriminant of the quadratic is: \[ D = b^2 - 4ac = 1^2 - 4(1)(4) = 1 - 16 = -15 \] Since \(D<0\), there are no real solutions in this interval. 
Case 3: \( x \geq -2 \)
In this interval, both \( x+4>0 \) and \( x+2 \geq 0 \), so the absolute values become: \[ |x+4| = x+4 \quad \text{and} \quad |x+2| = x+2 \] Substituting these into the equation: \[ x(x+4) + 3(x+2) + 10 = 0 \] Simplifying: \[ x^2 + 4x + 3x + 6 + 10 = 0 \] \[ x^2 + 7x + 16 = 0 \] The discriminant of the quadratic is: \[ D = b^2 - 4ac = 7^2 - 4(1)(16) = 49 - 64 = -15 \] Since \(D<0\), there are no real solutions in this interval. 
Step 4: Conclusion
After analyzing all three cases, we find only one real solution from Case 1: \[ x = \frac{-7 - \sqrt{65}}{2} \] Therefore, the equation has one real solution.