6x2 - x - 15 = 0 since (6)(-15) = (-10)(9) and -10 + 9 = -1, then factor as:
[(6x + 9)/3][6x - 10)/2] = 0
(2x + 3)(3x - 5) = 0 let each factor be zero:
2x + 3 = 0 or 3x - 5 = 0 solve for x:
The solutions are: x = -3/2 and x = 5/3
'x' must be zero, so the solution set is [ 0 ].
4a + 6 - 4a = -10 Combining like terms: 6 = -10 or 4 = 0 which is FALSE. So the solution set is empty.
It has infinitely many solutions.
Yes.
X2 - X - 6 = 0what two factors of - 6 add up to - 1 ?(X + 2)(X - 3)============(- 2, 0 ) and (3, 0 )------------------------------solution set of points
a = [ 0, 4 ]
x-2y=0 x=2y The solution set is the set of all (x,y) such that x=2y
x = 0 or x = 2
(0, -14)
'x' must be zero, so the solution set is [ 0 ].
(-2,-1)
17
Not sure about the set builder notation, but Q = {0}, the set consisting only of the number 0.
Using the quadratic formula, I found the solution set is x=2,x=-9
7 and (-3/2)
x2 + 49 = 0 ∴ x2 = -49 ∴ x = 7i
r=0 is the solution...