Question

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

Answer 1

J can be 2..

Answer 2

a = 1, b = 3 and c= j

Find D:

\[D = b^2 - 4ac\]

\[D = 3^2 - 4j\]

For real roots D must be greater than 0..

So, for what value of j is
\[9 - 4j > 0\]

That will be your answer..

Answer 3

This is a quadratic, and you can find its roots (solutions for x that make it true) using the quadratic formula
1 2 (−b±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

Check delta.
\[3^2 - 4(1)(j)\ge0\]\[9 - 4j\ge0\]\[9 \ge4j\]\[j\le \frac{9}{4}\]
''two real number solutions''... Is repeated counted as two or one?! Personally, I think it is counted as 2...

Answer 5

Thank you both