The number of real solution of equation x$|$x+4$|$+3$|$x+2$|$+10=0 is/are :
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- 1
- 2
- 3
- 4
The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
We need to find the number of real solutions for the equation involving absolute value functions: \(x|x+4|+3|x+2|+10=0\). To solve this, we must analyze the equation in different intervals based on the points where the expressions inside the absolute value signs become zero.
Step 2: Identifying Critical Points and Intervals:
The critical points are where the arguments of the absolute value functions are zero.
These are \(x+4=0 \Rightarrow x=-4\) and \(x+2=0 \Rightarrow x=-2\).
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: Solving the Equation in Each Interval:
Case 1: \(x<-4\)
In this interval, \(x+4<0\) and \(x+2<0\).
So, \(|x+4| = -(x+4)\) and \(|x+2| = -(x+2)\).
The equation becomes: \[ x(-(x+4)) + 3(-(x+2)) + 10 = 0 \] \[ -x^2 - 4x - 3x - 6 + 10 = 0 \] \[ -x^2 - 7x + 4 = 0 \] \[ x^2 + 7x - 4 = 0 \] Using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\): \[ x = \frac{-7 \pm \sqrt{7^2 - 4(1)(-4)}}{2(1)} = \frac{-7 \pm \sqrt{49 + 16}}{2} = \frac{-7 \pm \sqrt{65}}{2} \] The two possible values for x are \(x_1 = \frac{-7 + \sqrt{65}}{2}\) and \(x_2 = \frac{-7 - \sqrt{65}}{2}\).
Since \(\sqrt{64}=8\), \(\sqrt{65} \approx 8.06\).
\(x_1 \approx \frac{-7 + 8.06}{2} = \frac{1.06}{2} = 0.53\). This is not in the interval \(x<-4\).
\(x_2 \approx \frac{-7 - 8.06}{2} = \frac{-15.06}{2} = -7.53\). This is in the interval \(x<-4\).
So, we have one solution from this case: \(x = \frac{-7 - \sqrt{65}}{2}\).
Case 2: \(-4 \leq x<-2\)
In this interval, \(x+4 \geq 0\) and \(x+2<0\).
So, \(|x+4| = x+4\) and \(|x+2| = -(x+2)\).
The equation becomes: \[ x(x+4) + 3(-(x+2)) + 10 = 0 \] \[ x^2 + 4x - 3x - 6 + 10 = 0 \] \[ x^2 + x + 4 = 0 \] The discriminant 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, \(x+4>0\) and \(x+2 \geq 0\).
So, \(|x+4| = x+4\) and \(|x+2| = x+2\).
The equation becomes: \[ x(x+4) + 3(x+2) + 10 = 0 \] \[ x^2 + 4x + 3x + 6 + 10 = 0 \] \[ x^2 + 7x + 16 = 0 \] The discriminant 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: Final Answer:
Combining the results from all three cases, we find only one real solution, \(x = \frac{-7 - \sqrt{65}}{2}\).
Therefore, the number of real solutions is 1.
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