Question

For the equation x^2 + 3x + j = 0, find all the values of j such that the equation has two real number solutions. Show your work. help

Answer 1

For the equation to have two real solutions, in the equation:

ax^2+bx+x=0

b^2-4ac>0

Thus:

3^2-4*1*c>0
9-4c>0
-4c>-9
c<-9/-4
c>9/4

Remember when you divide an inequality by a divide sign you must flip the signs.

I think this is right, I could be wrong.

Answer 2

for an equation like:-\[ax^2+bx+c=0.\] u should know that it has 2 real solutions if D>0
I mean \[b^2-4ac>0\]
I t will give u j's value.
gt it:)
@letina

Answer 3

thus, J must be greater then 9/4.