2007-08-23 04:43:15 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
ALL quadratics have a maximum of 2 solutions. No exceptions.
x^2-x-6=0
(x-3)(x+2)=0
Now for (x-3)(x+2) to equal zero, at least one factor has to be zero
If x-3=0, then x=3
If x+2=0,then x=-2
That's all there is to it!
2007-08-23 04:50:51 · answer #1 · answered by Grampedo 7 · 1⤊ 0⤋
Factoring:
= x^2 - x - 6
= (x - 3) (x + 2)
Therefore:
x - 3 = 0, x + 2 = 0
x = 3, x = - 2
Answer: x = 3 and x = - 2
2007-08-27 07:31:42 · answer #2 · answered by Jun Agruda 7 · 3⤊ 0⤋
x^2 tells you there will be 2 solutions. x^3 would tell you there are a possible 3 solutions. x would have one solution.
factor and get (x-3)(x-2)=0, The two answers are obtained by setting both ( )'s equal to zero.
x-3=0, so x=3
x+2=0, so x=-2
The two answers are 3 and -2.
2007-08-23 11:52:49 · answer #3 · answered by Ed S 4 · 0⤊ 0⤋
solutions x=3 and x=-2 (check it is true)
2007-08-23 11:49:05 · answer #4 · answered by maussy 7 · 0⤊ 0⤋
(x - 3) (x + 2) = 0
x = 3 , x = - 2
2007-08-27 05:31:41 · answer #5 · answered by Como 7 · 0⤊ 0⤋