Complex Numbers And Quadratic Equations question 200
Question: Let $ a,b,c $ be real numbers $ a\ne 0 $ . If $ \alpha $ is a root $ a^{2}x^{2}+bx+c=0 $ , $ \beta $ is a root of $ a^{2}x^{2}-bx-c=0 $ and $ 0<\alpha <\beta $ , then the equation $ a^{2}x^{2}+2bx+2c=0 $ has a root $ \gamma $ that always satisfies [IIT 1989]
Options:
A) $ \gamma =\frac{\alpha +\beta }{2} $
B) $ \gamma =\alpha +\frac{\beta }{2} $
C) $ \gamma =\alpha $
D) $ \alpha <\gamma <\beta $
Show Answer
Answer:
Correct Answer: D
Solution:
Since $ \alpha $ and $ \beta $ are the roots of given equations. So we have $ a^{2}{{\alpha }^{2}}+b\alpha +c=0 $ and $ a^{2}{{\beta }^{2}}-b\beta -c=0 $ . Let $ f(x)=a^{2}x^{2}+2bx+2c=0 $ Then $ f(\alpha )=a^{2}{{\alpha }^{2}}+2b\alpha +2c=0 $ $ =a^{2}{{\alpha }^{2}}+2(b\alpha +c)=a^{2}{{\alpha }^{2}}-2a^{2}{{\alpha }^{2}}=-a^{2}{{\alpha }^{2}}=-ve $ and $ f(\beta )=a^{2}{{\beta }^{2}}+2(b\beta +c)=a^{2}{{\beta }^{2}}+2a^{2}{{\beta }^{2}} $ $ =3a^{2}{{\beta }^{2}}=+ve $ Since $ f(\alpha ) $ and $ f(\beta ) $ are of opposite signs, therefore by theory of equations there lies a root $ \gamma $ of the equation $ f(x)=0 $ between $ \alpha $ and $ \beta $ i.e. $ \alpha <\gamma <\beta $
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