Question

Find the value of p for which the quadratic equation 2x²+px+8=0 has real roots.​

Answer 1

Answer:

$b + (\sqrt{ {b}^{2} } - \sqrt{4ac }) \div 2a$

Step-by-step explanation:

a=2, b=p, c=8

{p+√(p)2-√4×2×8}÷ 2×2=0

{p+(p)2-64}÷4=0

(p+p-8)÷4=0

2p-8=0×4

2p-8=0

p=8÷2

p=4