Question 135087: Hello,I was wondering if anyone could help me solve this equation? I'm having trouble with it.
x^2-3x=18 the directions are to solve the equation.
Thank you,
Mrs. Gibson
Answer by algebrapro18(249)   (Show Source):
You can put this solution on YOUR website! Well in order to solve a quadratic equation we first must have it in standard form. Standard form is ax^2+bx+c = 0 where a, b, and c are numbers.
so putting the function into standard form you get:
x^2-3x=18
x^2-3x-18=0
now we need to factor. The steps to factoring are:
1) Multiply a times c
2) See what factors of the number found in step one add to the b value
3) break the function into a product of its factors
In this example a = 1, b = -3, and c = -18. So a*c = -18. So we need to find the factors of -18 that add to -3. Well the factors of 18 are:
1,-18
2,-9
3,-6
well -6+3 = -3 so those are the factors we wanted.
So we now know that x^2-3x-18 can be written as (x-6)(x+3). This is now in factored form.
Now we use the zero product property that states that you can set each one of those factors equal to zero and solve for x.
x-6 = 0
x = 6
x+3 = 0
x = -3
so now we know that the solutions to the given equation are x = -3 and x = 6.
Now to check our solutions:
x^2-3x-18 = 0
(6)^2-3(6)-18=0
36 - 18 - 18 = 0
36 - 36 = 0
0 = 0
that solution checks out now lets check the other one
x^2-3x-18 = 0
(-3)^2 - 3(-3) - 18 = 0
9 + 9 - 18 = 0
18 - 18 = 0
0 = 0
that solution also check's out.
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