Question

find the value of p which quadratic equation (2p +1)x2 -(7p+2)x +(7p-3) =0 has equal roots. also find these roots

Answer 1

AS roots are equal,
Means, the situation should be,
b square minus 4ac=0
Therefore,
(7p+2)square -4.(2p+1)=0
49p square +28p+4-8p-4=0
49p square+20p=0
P(49p+20)=0
49p+20==0
Therefore p=-20/49

Answer 2

ANSWER

Given Eqn is  

a

(2p+1)x  

2

 

​  

−  

b

(7p+2)x

​  

+  

c

7p−3

​  

=02

If this eqn has equal roots Discriminant

D=0

b  

2

−4ac=0

⇒(7p+2)  

2

−4(2p+1)(7p−3)=0

⇒49p  

2

+4+28p−4(149  

2

−6p−3+7p)=0

⇒49p  

2

+4+28p−(56p  

2

−24p−12+28p)=0

⇒49p  

2

+4+28p−56p  

2

−4p+12=0

⇒−7p  

2

+24p+16=0

⇒7p  

2

−24p−16=0⇒P=4,  

7

−4

​  

 

Now Roots At

i)P=4                                                ii)P=  

7

−4

​  

 

Eqn is 9x  

2

−30x+25=0               Eqn is  

7

−1

​  

x  

2

+2x−7=0

3x−5=0                                             x  

2

−14x+49=0

x=  

3

5

​  

                                                       x=7