Question

Hi guys, can someone please help with this assignment :

For which value of p with the equation below have no real solution


px^2 + 4x + 6 = 0

I got so far, but stuck and not sure where to go further:


px^2 + 4x + 6 = 0 >> I divide then everything with p


x^2 + 4x/p + 6/p = 0


x= -4/2p +- √(-4/2p)^2 -6/p

x= -2/p +-√4/p -6/p


I'm not sure where to go from here. Can someone please help me.

Thanks in advanced for any help!

Answer 1

p >
`(2)/(3)`

given a quadratic equation in standard form : ax² + bx + c = 0 : a ≠ 0

We can determine the nature of the roots using the discriminant

Δ = b² - 4ac

• If b² - 4ac > 0 , then 2 real and distinct roots

• If b² - 4ac = 0, then 2 real and equal roots

• If b² - 4ac < 0, then no real roots

px² + 4x + 6 = 0 is in standard form

with a = p, b = 4 and c = 6

For no real solution, solve b² - 4ac < 0

4² - (4 × p × 6 ) < 0

16 - 24p < 0 ( subtract 16 from both sides )

- 24p < - 16 ( divide both sides by - 16 , reversing the symbol in doing so )

p >
`(16)/(24)`

Thus for the equation to have no real solutions p >
`(2)/(3)`