Complex Numbers And Quadratic Equations question 772
Question: If the roots of the equation $ x^{2}-8x+(a^{2}-6a)=0 $ are real, then [RPET 1987, 97; MP PET 1999]
Options:
A) $ -2<a<8 $
B) $ 2<a<8 $
C) $ -2\le a\le 8 $
D) $ 2\le a\le 8 $
Show Answer
Answer:
Correct Answer: C
Solution:
Roots of $ x^{2}-8x+(a^{2}-6a)=0 $ are real. So $ D\ge 0 $
Þ $ 64-4(a^{2}-6a)\ge 0 $
Þ $ 16-a^{2}+6a\ge 0 $
Þ $ a^{2}-6a-16\le 0 $
Þ $ (a-8)(a+2)\le 0 $ Now we have two cases: Case I : $ (a-8)\le 0 $ and $ (a+2)\ge 0 $
Þ $ a\le 8 $ and $ a\ge -2 $ Case II : $ (a-8)\ge 0 $ and $ (a+2)\le 0 $
Þ $ a\ge 8 $ and $ a\le -2 $ but it is impossible Therefore, we get $ -2\le a\le 8 $ Aliter : Students should note that the expression $ (x-a)(x-b){a<b} $ will be less than or equal to zero if $ x\in [a,b] $ or otherwise $ x\notin [a,b] $ . Therefore $ (a-8)(a+2)\le 0 $ i.e., $ {a-(-2)}(a-8)\le 0\Rightarrow a\in [-2,\ 8] $ .
BETA
