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

if it has two real number solutions that means \[b^2 - 4ac > 0\]
so substitute
\[(3)^2 + 4(1)(j) > 0\]
\[\implies 9 + 4j > 0\]
subtract 9 from both sides
\[\implies 4j > -9\]
divide both sides by 4
\[\implies j > -\frac 94\]

Answer 2

tell me if you have questions about the solution

Answer 3

if you have no idea what i did that was the "discriminant rule"
\(b^2 - 4ac>0\) <--2 real solutions
\(b^2 - 4ac = 0\) <--one real solution
\(b^2 - 4ac < 0\) <--no real solution