Question

find the values of p for which the roots are real and equal P^2 x^2 - 2 (2p- 1) x+4 =0 the answer is 1 /4 give the solution​

Answer 1

Explanation:

Given Eqn is a(2p+1)x2−b(7p+2)x+c7p−3=02

If this eqn has equal roots Discriminant

D=0

b2−4ac=0

⇒(7p+2)2−4(2p+1)(7p−3)=0

⇒49p2+4+28p−4(1492−6p−3+7p)=0

⇒49p2+4+28p−(56p2−24p−12+28p)=0

⇒49p2+4+28p−56p2−4p+12=0

⇒−7p2+24p+16=0

⇒7p2−24p−16=0⇒P=4,7−4

Now Roots At

i)P=4                                                ii)P=