Complex Numbers And Quadratic Equations question 639
Question: If one root of the equation $ (1-m)x^{2}+1x+1=0 $ is double the other and 1 is real, then what is the greatest value of m?
Options:
A) $ -\frac{9}{8} $
B) $ \frac{9}{8} $
C) $ -\frac{8}{9} $
D) $ \frac{8}{9} $
Show Answer
Answer:
Correct Answer: B
Solution:
Given equation is $ (\ell -m)x^{2}+\ell x+1=0 $ Roots are $ \alpha ,\beta . $ $ \because $ One root is double the other. $ \beta =2\alpha $ Sum of roots $ =\alpha +\beta $ $ 3\alpha =\frac{-\ell }{\ell -m} $ $ \alpha (2\alpha )=\frac{1}{(\ell -m)} $
$ \Rightarrow {{\alpha }^{2}}=\frac{{{\ell }^{2}}}{9{{(\ell -m)}^{2}}} $ $ 2{{\alpha }^{2}}=\frac{1}{\ell -m} $
$ \Rightarrow 2\frac{{{\ell }^{2}}}{9{{(\ell -m)}^{2}}}=\frac{1}{(\ell -m)} $
$ \Rightarrow \frac{2{{\ell }^{2}}}{9(\ell -m)}=1 $
$ \Rightarrow 2{{\ell }^{2}}=9(\ell -m)\Rightarrow 2{{\ell }^{2}}-9\ell +9m=0 $ For $ \ell $ to be real discriminant should be $ b^{2}-4ac\ge 0 $ $ 81-4\times 2\times 9m\ge 0 $ $ m\le \frac{9}{8}. $
BETA
