JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations
The number of distinct real solutions of the equation \(x|x+4|+3|x+2|+10=0\) is
- A 3
- B 1
- C 0
- D 2
Answer & Solution
Correct Answer
(B) 1
Step-by-step Solution
Detailed explanation
Case I \(x < -4\) \(x(-(x+4)) + 3(-(x+2)) + 10 = 0\) \(x^2+7 x-4=0\) \(\underset {\text {Reject}} {x=-\frac{7+\sqrt{65}}{2}}\)or \(\underset {\text{Acept}} {x=-\frac{7-\sqrt{65}}{2}}\) Case II \(-4 \leq x<-2\) \(x(x+4)+3(-(x+2))+10=0\) \(x^2+x+4=0\) \(D<0\)\(\quad\)No solution…
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