Complex Numbers And Quadratic Equations question 180
Question: If the roots of the equations $ x^{2}-bx+c=0 $ and $ x^{2}-cx+b=0 $ differ by the same quantity, then $ b+c $ is equal to [BIT Ranchi 1969; MP PET 1993]
Options:
A) 4
B) 1
C) 0
D) -4
Show Answer
Answer:
Correct Answer: D
Solution:
Let the roots are $ \alpha ,\beta $ of $ x^{2}-bx+c=0 $ and $ {\alpha }’,{\beta }’ $ be roots of $ x^{2}-cx+b=0 $ Now $ \alpha -\beta =\sqrt{{{(\alpha +\beta )}^{2}}-4\alpha \beta }=\sqrt{b^{2}-4c} $ …..(i) and $ \alpha ‘-\beta ‘=\sqrt{{{(\alpha ‘+\beta ‘)}^{2}}-4\alpha ‘\beta ‘}=\sqrt{c^{2}-4b} $ …..(ii) But $ \alpha -\beta =\alpha ‘-\beta ’ $
Þ $ \sqrt{b^{2}-4c}=\sqrt{c^{2}-4b}\Rightarrow b^{2}-4c=c^{2}-4b $
Þ $ b^{2}-c^{2}=4c-4b $
Þ $ (b+c)(b-c)=4(c-b) $
Þ $ b+c=-4 $
BETA
