Complex Numbers And Quadratic Equations question 184
Question: If one root of the equation $ ax^{2}+bx+c=0 $ the square of the other, then $ a{{(c-b)}^{3}}=cX $ , where X is
Options:
A) $ a^{3}+b^{3} $
B) $ {{(a-b)}^{3}} $
C) $ a^{3}-b^{3} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
If one root is square of other of the equation $ ax^{2}+bx+c=0 $ , then $ b^{3}+ac^{2}+a^{2}c=3abc $ Which can be written in the form $ a{{(c-b)}^{3}}=c{{(a-b)}^{3}} $ Trick: Let roots be 2 and 4, then the equation is $ x^{2}-6x+8=0 $ . Here obviously $ X=\frac{a{{(c-b)}^{3}}}{c}=\frac{1{{(14)}^{3}}}{8}=\frac{14}{2}\times \frac{14}{2}\times \frac{14}{2}=7^{3} $ Which is given by $ {{(a-b)}^{3}}=7^{3} $ .
BETA
