Question

find p, if root are equal (2p-3)x^2+8x+1=0​

Answer 1

Answer:

Given equation: (2p-3)$x^{2}$ + 8x + 1=0​

a= (2p -3)

b= 8

c=1

It is given that the roots of the equation are equal.

Therefore,

$b^{2}$- 4ac = 0

Putting the values, we get:

$=> 8^{2} - 4 ( 2p -3) (1) = 0 \\=>64 - 8p +12 =0 \\=> 4 ( 16 - 2p +3 ) = 0 \\=> -2p +16 +3 =0 \\=> -2p + 19 =0 \\=> -2p =-19 \\=> p = -19/-2\\=> p= 19/2$

Hence, when the roots of the given equation are equal, the value of p = 19/2