SOLUTION: Solve the equation. 1. 3a^2=-a+2 2. 3(x-3)=2(6-x) I am having alot of trouble with these three problems. I got x=-9 for problem two. I need for somebody to show me

Algebra ->  Coordinate Systems and Linear Equations  -> Lessons -> SOLUTION: Solve the equation. 1. 3a^2=-a+2 2. 3(x-3)=2(6-x) I am having alot of trouble with these three problems. I got x=-9 for problem two. I need for somebody to show me       Log On


   

Question 32822: Solve the equation.
1. 3a^2=-a+2
2. 3(x-3)=2(6-x)

I am having alot of trouble with these three problems. I got x=-9 for problem two. I need for somebody to show me the steps so that I can understand them for the test. These questions do not come from a book they are comming from a review sheet. I know there is only one submission per day but I really need help on these three problms. Thanks hope you can help.
Found 2 solutions by stanbon, Nate:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
1. 3a^2+a-2=0
Use the quadratic formula as illustrated below:
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation aa%5E2%2Bba%2Bc=0 (in our case 3a%5E2%2B1a%2B-2+=+0) has the following solutons:

a%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%281%29%5E2-4%2A3%2A-2=25.

Discriminant d=25 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-1%2B-sqrt%28+25+%29%29%2F2%5Ca.

a%5B1%5D+=+%28-%281%29%2Bsqrt%28+25+%29%29%2F2%5C3+=+0.666666666666667
a%5B2%5D+=+%28-%281%29-sqrt%28+25+%29%29%2F2%5C3+=+-1

Quadratic expression 3a%5E2%2B1a%2B-2 can be factored:
3a%5E2%2B1a%2B-2+=+3%28a-0.666666666666667%29%2A%28a--1%29
Again, the answer is: 0.666666666666667, -1. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+3%2Ax%5E2%2B1%2Ax%2B-2+%29

2.You will need to post this with parentheses properly
placed so we can tell where numerators and denominators
start and end.
3. Expanding the original problem you get the following:
3x-9=12-2x
3x-2x=12+9
x = 21
Cheers,
Stan H.

Answer by Nate(3500) About Me  (Show Source):
You can put this solution on YOUR website!
1. 3a^2=-a+2
first, let us get the whole equation equal to zero; so:
3a^2+a-2=0
then, multiply the leading coefficient(the coefficient with the variable that has the highest degree; it is 3 in this problem) with the constant (the number that doesn't have a variable; it is -2 in this problem) to get -6......
now, find factors of -6 that add up to the coefficient of the middle term (1a)
Those factors are 3 and -2....so let us fit those numbers into the problem
3a^2+a-2=0 original
3a^2+3a-2a-2=0 original with factors
(3a^2+3a)+(-2a-2)=0 grouping
3a(a+1)-2(a+1)=0 take out the common multiple
(3a-2)(a+1)=0 combine
now, use the zero property to find what (a) is:
3a-2=0
a=2/3
--------------
a+1=0
a=-1
SO, -1 and 2/3 are your answers!
-----------------------------------------------------------------------
2. 3(x-3)=2(6-x)
3(x-3)=2(6-x) original
3x-9=12-2x distribute
5x=21 get all variables to one side and vise versa with constant numbers
x=21/5 your answer!