logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

If $\;\large\frac{1}{\sqrt{\alpha}}\;$ and $\;\large\frac{1}{\sqrt{\beta}}\;$ are the roots of the equation ,$\;ax^{2}+bx+c=0 \;(a \neq 0 , q ,b \in R)\;$, then the equation , $\;x(x+b^{3}) + (a^{3}-3abx)=0\;$ has roots :

$(a)\;\alpha^{\large\frac{3}{2}}\;and\; \beta^{\large\frac{3}{2}}\qquad(b)\;\alpha \beta^{\large\frac{3}{2}}\;and\; \beta \alpha^{\large\frac{3}{2}}\qquad(c)\;\sqrt{\alpha \beta} \;and\; \alpha \beta \qquad(d)\;\alpha^{-\large\frac{3}{2}}\;and\; \beta^{-\large\frac{3}{2}}$

Can you answer this question?
 
 

1 Answer

0 votes
The question is wrong, the correct equation is, ax^2+bx+1=0   , as the second equation involves no c.

Write the equation as x^2 + (b^3-3ab)x+a^3=0    

The product of root of this equation is   [(âß)]^3/2   , which is only satiesfied by (a) optiion.
answered Apr 2, 2015 by Limitsofawesomeness
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...