Question

For the equation x2 + 3x + j = 0, find all the values of j such that the equation has two real number solutions. Show your work.

Answer 1

you need \(b^2-4ac>0\) so in this case you must solve \(9-4\times j>0\)

Answer 2

for 2 real roots 3^2 - 4*j > 0

Answer 3

Δ = 3^2 - 4j
= 9-4j

the equation has two real number solutions for Δ>0

so 9-4j>0
j< 9/4