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.
To find the values of j such that the equation has two real number solutions, we need to consider the discriminant of the quadratic equation.
The discriminant (D) is given by the formula D = b^2 - 4ac, where a, b, and c are the coefficients of the quadratic equation ax^2 + bx + c = 0.
In this case, the equation is x^2 + 3x + j = 0, so a = 1, b = 3, and c = j. Plugging these values into the discriminant formula, we get:
D = (3)^2 - 4(1)(j)
D = 9 - 4j
For the equation to have two real number solutions, the discriminant must be greater than zero. Mathematically, this condition is represented as D > 0.
Therefore, we have:
9 - 4j > 0
To solve for j, we can subtract 9 from both sides:
-4j > -9
Now, divide both sides by -4, but remember to reverse the inequality sign since we are dividing by a negative number:
j < 9/4
Hence, all values of j that satisfy the inequality j < 9/4 will give the quadratic equation x^2 + 3x + j = 0 two real number solutions.