Question

Find the values of p for which the equadratic equation (2p+1) x2-(7p+2)x+(7p-3)=0
has equal roots. Also, find these roots.
[CBSE 2014)​

Answer 1

Answer:

ax^2+bx+c=0 in this quadratic equation the condition of two equal roots is b^2-4ac=0

here a=(2p+1)

b= -(7p+2)

c=(7p-3)

so,

{-(7p+2)}^2-4(2p+1)(7p-3)=0

=> 49p^2+4+28p-56p^2-28p+24p+12=0

=> -7p^2+24p+16=0

=> -7p^2+28p-4p+16=0

=> -7p(p-4)-4(p-4)=0

=> (p-4)(-7p-4)=0

so,

p=4

so the equation is

9x^2-30x+25=0

=> (3x)^2-2.3x.5+(5)^2=0

=> (3x-5)^2=0

=> (3x-5)=0

=> 3x=5

=> x=5/3=1 2/3

the value of p=4

the roots are 1 2/3 and 1 2/3