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