3x^2+bx+12=0 b? The equation must have two same solutions
The discriminant must = 0 if this is true
So b^2 - 4(3)(12) = 0
b^2 = 144
b = ±12 [ -2 and 2 will be solutions.....each with a multiplicity of 2 ]
x^2+bx-24=0 b? The solution must be -3
We must have this
(-3)^2 + b(-3) - 24 = 0
9 - 3b - 24 = 0
-15 - 3b = 0
-15 = 3b
b = -5 [ 8 will also be a solution ]
x^2-mx+8=0 m? The solution must be 4
(4)^2 - m(4) + 8 = 0
16 - 4m + 8 = 0
24 - 4m = 0
24 = 4m
6 = m [ 2 will also be a solution ]
Write an equation with solutions -2 and -17
P(x) = (x + 2) (x + `17) = x^2 + 19x + 34
Write an equation with solutions -11 and 9, and the coefficient of x^2 must be -4
We have
P(x) = -4 (x + 11) (x - 9) = -4 [x^2 + 2x - 99] = -4x^2 - 8x + 396
