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

Look for all j such that the determinant is >= 0, by using the general quadratic solution. a=1, b=3, and c=j.

Answer 2

General quadratic solution is x = [-b +- sqrt(b^2 - 4ac)] / (2a). To find real solutions, you only have to concern yourself with b^2 - 4ac and make sure that is >= 0.

Answer 3

I dont know how to solce b^2 - 4ac

Answer 4

Substitute 3 for b, 1 for a, and j for c, and set it >= 0. There, that's all you have left to do. 90% done for you. Just make the substitutions.

Answer 5

j less than or equal to 0

Answer 6

That's not the answer. Show me your work and then I can tell you where you went wrong.

Answer 7

idk how to show work...im horrible at math

Answer 8

9-4ac

Answer 9

You correctly substituted for the b. Continue substituting the a and the c. I gave you the values for a, b, and c in my first post.

Answer 10

i dont know how to do the rest :( thats why i stopped

Answer 11

This might help. You wrote 9-4ac and that is correct and you have to set that >= 0, so you have 9 - 4ac >= 0 This is what should help: 9 - 4ac >= 0 is the same as 9 - (4)(a)(c) >= 0. Now, even if you can't solve the equation, start by putting the values I gave you in place of the a and the c.

Answer 12

What is "a"? What is "c"? Are you still there?

Answer 13

@tcarrol010, 1 for a, and j for c,