logo

Ask Questions, Get Answers

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

$\alpha,\beta$ are roots of $ax^2+bx+c=0$.The value of $(a\alpha+b)^{-2}+(a\beta+b)^{-2}$ is

$\begin{array}{1 1}(A)\;\large\frac{b^2-2ac}{a^2c^2}&(B)\;\large\frac{c^2-2ab}{a^2b^2}\\(C)\;\large\frac{a^2-2bc}{b^2c^2}&(D)\;\text{None}\end{array} $

Can you answer this question?
 
 

1 Answer

0 votes
$(a\alpha+b)^{-2}+(a\beta+b)^{-2}$
$\Rightarrow a^{-2}\big[(\alpha+\large\frac{b}{a})^{-2}+(\beta+\large\frac{b}{a})^{-2}\big]$
$\alpha+\beta=-\large\frac{b}{a}$$\Rightarrow \large\frac{b}{a}$$=-\alpha-\beta$
$\Rightarrow a^{-2}\big[(\alpha+\large\frac{b}{a}^{-2}+$$(\beta+\large\frac{b}{a})^{-2}\big]$
$\Rightarrow a^{-2}\big[(\alpha-\alpha-\beta)^{-2}+(\beta-\alpha-\beta)^{-2}\big]$
$\Rightarrow a^{-2}\big[\beta^{-2}+\alpha^{-2}\big]$
$\Rightarrow \large\frac{1}{a^2}\frac{(\alpha^2+\beta^2)}{\alpha^2\beta^2}$
$\Rightarrow \large\frac{b^2-2ac}{a^2c^2}$
Hence (A) is the correct answer.
answered Apr 15, 2014 by sreemathi.v
 

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
...