Complex Numbers And Quadratic Equations question 788
Question: The roots of the equation $ (a^{2}+b^{2})t^{2}-2(ac+bd)t+(c^{2}+d^{2})=0 $ are equal, then [MP PET 1996]
Options:
A) $ ab=dc $
B) $ ac=bd $
C) $ ad+bc=0 $
D) $ \frac{a}{b}=\frac{c}{d} $
Show Answer
Answer:
Correct Answer: D
Solution:
Accordingly, $ {{{2(ac+bd)}}^{2}}=4(a^{2}+b^{2})(c^{2}+d^{2}) $
Þ $ 4a^{2}c^{2}+4b^{2}d^{2}+8abcd=4a^{2}c^{2}+4a^{2}d^{2} $ $ +4b^{2}c^{2}+4b^{2}d^{2} $
Þ $ 4a^{2}d^{2}+4b^{2}c^{2}-8abcd=0\Rightarrow 4{{(ad-bc)}^{2}}=0 $
Þ $ ad=bc\Rightarrow \frac{a}{b}=\frac{c}{d} $ .
BETA
