did you get
x = (15 ± √(225 - 4(3)(18) )/6 ?
= (15 ± √9)/6
= (15 ±3)/6
= 3 or 2
we could have divided the original equation by 3 to get
x^2 - 5x + 6 = 0
now it factors
(x-3)(x-2) = 0
so x = 3 or x = 2
Solve the quadratic equation.
3x^2-15x+18=0
I know that the formula to quadratic equation is x=-b+and-(square root)b^2-4ac divided by 2a.
I know that a=3, b=-15 and c=18, then subsitute in and get two answers, my problem is that I keep getting an answer that does not work after I subsitute in the original equation. What do I do?
3 answers
I got 3 and -2. But to check it, wouldn't you substitue it the 3 and then the -2 in the original equation and it should be 0=0?
and so it does
How did you get -2?
if you had a + and a - then you couldn't possible end up with +18 at the end.
if x= 3
3(9) - 45 + 18 = ??
if x=2
3(42) - 30 + 18 = ???
How did you get -2?
if you had a + and a - then you couldn't possible end up with +18 at the end.
if x= 3
3(9) - 45 + 18 = ??
if x=2
3(42) - 30 + 18 = ???
1 answer