Complex Numbers And Quadratic Equations question 167
Question: If $ ax^{2}+bx+c=0 $ , then x = [MP PET 1995]
Options:
A) $ \frac{b\pm \sqrt{b^{2}-4ac}}{2a} $
B) $ \frac{-b\pm \sqrt{b^{2}-ac}}{2a} $
C) $ \frac{2c}{-b\pm \sqrt{b^{2}-4ac}} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a} $ Since $ \frac{2c}{-b+\sqrt{b^{2}-4ac}}.\frac{-b-\sqrt{b^{2}-4ac}}{-b-\sqrt{b^{2}-4ac}} $ = $ \frac{2c(-b-\sqrt{b^{2}-4ac})}{4ac}=\frac{-b-\sqrt{b^{2}-4ac}}{2a} $ Similarly $ \frac{2c}{-b-\sqrt{b^{2}-4ac}}\times \frac{-b+\sqrt{b^{2}-4ac}}{-b+\sqrt{b^{2}-4ac}} $ = $ \frac{2c(-b+\sqrt{b^{2}-4ac})}{4ac}=\frac{-b+\sqrt{b^{2}-4ac}}{2a} $ Aliter : On rationalising the given equation $ x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a} $ , we get option C correct.
BETA
