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

@phi

Answer 2

fds.oup.com/www.oup.com/pdf/13/9780199117796.pdf

Answer 3

I assume you mean
x^2 + 3x + j=0

This is a quadratic, and you can find its roots (solutions for x that make it true) using the quadratic formula
\( \frac{1}{2}(-b±\sqrt{b^2-4ac}) \) where we match up the a,b,c with \(ax^2+bx+c\)
You get real solutions when the stuff inside the square root is 0 or positive. People call the expression \(b^2-4ac \) the DISCRIMINATE . So find all values of j (c in the equation) that make the discriminate ≥0

Answer 4

*DISCRIMINANT