Question

Find the value of p for which the quadratic equation `2x^2+px+8=0` has real roots. 

Answer 1

Given: 

`2x^2+px+8` 

Here, 

`a=2, b=p and c=8` 

Discriminant D is given by: 

`D=(b^-4ac)` 

=`p^2-4xx2xx8` 

=`(p^2-64)`  

If D > 0, the roots of the equation will be real 

⇒`(p^2-64)≥0 ` 

⇒` (p+8) (p-8)≥0` 

⇒ `p≥ 8  and p ≤-8` 

Thus, the roots of the equation are real for `p≥ 8` and `p ≤-8`