Complex Numbers And Quadratic Equations question 728
Question: The number of real solutions of the equation | $ x^{2} $ + 4x + 3| + 2x + 5 = 0 are [IIT 1988]
Options:
A) 1
B) 2
C) 3
D) 4
Show Answer
Answer:
Correct Answer: B
Solution:
Here two cases arise viz. Case I : $ x^{2}+4x+3>0 $ This gives $ x^{2}+4x+3+2x+5=0 $
Þ $ x^{2}+6x+8=0 $
Þ $ (x+2)(x+4)=0 $
Þ $ x=-2,-4 $ $ x=-2 $ is not satisfying the condition $ x^{2}+4x+3>0 $ , so $ x=-4 $ is the only solution of the given equation. Case II : $ x^{2}+4x+3<0 $ This gives - $ (x^{2}+4x+3)+2x+5=0 $
Þ $ -x^{2}-2x+2=0\Rightarrow x^{2}+2x-2=0 $
Þ $ (x+1+\sqrt{3})(x+1-\sqrt{3})=0 $
Þ $ x=-1+\sqrt{3},-1-\sqrt{3} $ Hence $ x=-(1+\sqrt{3}) $ satisfy the given condition $ x^{2}+4x+3<0 $ , while $ x=-1+\sqrt{3} $ is not satisfying the condition. Thus number of real solutions are two.
BETA
