Question

Find the value of ‘p’ for which the roots of the following equation are real and equal:

(3p + 1) x² + 2(p + 1) x + p = 0

Answer 1

Given:

(3p + 1) x² + 2(p + 1) x + p = 0

For a quadratic equation ax² + bx + c = 0, the discriminant is:

D = b² − 4ac

For real and equal roots, we set:

D = 0

From the given equation:

a = 3p + 1

b = 2(p + 1)

c = p

D = [2(p + 1)]² − 4(3p + 1) (p)

[2(p + 1)]²

= 4(p + 1)²

= 4(p² + 2p + 1)

4(3p + 1) (p)

= 4p(3p + 1)

= 12p² + 4p

D = 4(p² + 2p + 1) − (12p² + 4p)

D = 4p² + 8p + 4 − 12p² − 4p

D = (4p² − 12p²) + (8p − 4p) + 4

= −8p² + 4p + 4

−8p² + 4p + 4 = 0

Multiply through by −1 to simplify:

8p² − 4p − 4 = 0

`p = (-(-4) +- sqrt((-4)^2 - 4(8)(-4)))/(2(8))`

`= (4+- sqrt(16+128))/16`

`= (4+-sqrt144)/16`

`= (4+-12)/16`

`p = (4+12)/16`

`= 16/16`

= 1

`p = (4-12)/16`

`=(-8)/16`

p = 1 or p = `-1/2`