Question

Find the value(s) of p for which the quadratic equation given as (p + 4) x² – (p + 1) x + 1 = 0 has real and equal roots. Also, find the roots of the equation(s) so obtained.

Answer 1

Given: The quadratic (p + 4) x² – (p + 1) x + 1 = 0.

Step-wise calculation:

1. For real and equal roots the discriminant D must be 0:

D = b² – 4ac, where a = p + 4, b = –(p + 1), c = 1.

2. Compute D: D = (–(p + 1))² – 4(p + 4)(1) 

= (p + 1)² – 4(p + 4) 

= p² + 2p + 1 – 4p – 16

= p² – 2p – 15

3. Set D = 0 and solve:

p² – 2p – 15 = 0 

(p – 5)(p + 3) = 0 

⇒ p = 5 or p = –3

4. For equal roots the repeated root is `x = -b/(2a)`. 

Since b = –(p + 1), –b = p + 1, so `x = (p + 1)/[2(p + 4)]`. 

Evaluate for each p:

If p = 5: `x = (5 + 1)/[2(5 + 4)]` 

= `6/18` 

= `1/3`

If p = –3: `x = (-3 + 1)/[2(-3 + 4)]` 

= `(-2)/2` 

= –1

The values of p giving real and equal roots are p = 5 and p = –3.

For p = 5 the (double) root is `x = 1/3`.

For p = –3 the (double) root is x = –1.