Hi guys, can someone 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 me. in advanced for any !

p >

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 >

Thus for the equation to have no real solutions p >

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation:

answer: it's d

step-by-step explanation: