Complex Numbers And Quadratic Equations question 809
Question: If the sum of the two roots of the equation $ 4x^{3}+16x^{2}-9x-36=0 $ is zero, then the roots are [MP PET 1986]
Options:
A) 1, 2 -2
B) $ -2,\frac{2}{3},-\frac{2}{3} $
C) $ -3,\frac{3}{2},-\frac{3}{2} $
D) $ -4,\frac{3}{2},-\frac{3}{2} $
Show Answer
Answer:
Correct Answer: D
Solution:
Given equation $ 4x^{3}+16x^{2}-9x-36=0 $ , Putting $ x=-4 $
Þ $ -4\times 64+256+36-36=0 $ Hence $ x=-4 $ is a root of the equation Now reduced equation is $ 4x^{2}(x+4)-9(x+4)=0 $
Þ $ (x+4)(4x^{2}-9)=0 $
Þ $ x=-4,x=\pm \frac{3}{2} $ Thus roots are $ -4,-\frac{3}{2},\frac{3}{2} $
BETA
