Complex Numbers And Quadratic Equations question 801
Question: The number of real values of x for which the equality $ | 3x^{2}+12x+6 |=5x+16 $ holds good is [AMU 1999]
Options:
A) 4
B) 3
C) 2
D) 1
Show Answer
Answer:
Correct Answer: C
Solution:
Equation is $ |3x^{2}+12x+6|=5x+16 $ …….(i) when $ 3x^{2}+12x+6\ge 0 $ $ \Leftrightarrow x^{2}+4x\ge -2 $ $ \Leftrightarrow |x+2{{|}^{2}}\ge 4-2\Leftrightarrow |x+2|\ge {{(\sqrt{2})}^{2}} $ $ \Leftrightarrow x+2\le -\sqrt{2} $ or $ x+2\ge \sqrt{2} $ ….(ii) Then (i) becomes $ 3x^{2}+12x+6=5x+16 $ $ \Leftrightarrow 3x^{2}+7x-10=0\Rightarrow x=1,-\frac{10}{3} $ But $ x=-\frac{10}{3} $ does not satisfy (ii). When $ 3x^{2}+12x+6<0 $
$ \Rightarrow x^{2}+4x<-2 $
Þ $ |x+2|\le \sqrt{2} $
$ \Rightarrow -\sqrt{2}-2\le x\le -2+\sqrt{2} $ …….(iii) Then (i) becomes
$ \Rightarrow 3x^{2}+12x+6=-(5x+16) $
$ \Rightarrow 3x^{2}+17x+22=0\Rightarrow x=-2,-\frac{11}{3} $ But $ x=-\frac{11}{3} $ does not satisfy (iii). So, 1 and - 2 are the only solutions.
BETA
