Question

Find the value of p for which the quadratic equation `(2p+1)x^2-(7p+2)x+(7p-3)=0` has real and equal roots. 

Answer 1

The given equation is `(2p+1)x^2-(7p+2)x+(7p-3)=0` 

This is of the form `ax^2+bx+c=0` where a=`2p+1, b=-(7p-2) and c=7 p-3` 

∴ `D=b^2-4ac` 

=`-[-7p+2]^2-4xx(2p+1)xx(7p-3)` 

=`(49p^2+28p+4)-4 (14p^2+p-3)` 

=`49p^2+28p+4-56p^2-4p+12` 

=`-7p^2+24p+16` 

The given equation will have real and equal roots if D = 0. 

∴ `-7p^2+24p+16=0`  

⇒ `7p^2-24p-16=0` 

⇒ `7p^2-28p+4p-16=0` 

⇒`7p(p-4)+4(p-4)=0` 

⇒`(p-4) (7p+4)=0` 

⇒` p-4=0 or 7p+4=0`  

⇒ `p=4 or p=-4/7` 

Hence, 4 and `-4/7` are the required values of p.