SOLUTION: Any help with these three would be appreciated. I keep thinking I've done them correctly but was actually wrong and I've gone around with them so much I think I've completely forg

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Any help with these three would be appreciated. I keep thinking I've done them correctly but was actually wrong and I've gone around with them so much I think I've completely forg      Log On


   



Question 184824: Any help with these three would be appreciated. I keep thinking I've done them correctly but was actually wrong and I've gone around with them so much I think I've completely forgotten what I'm doing!
1. 3x2 + 11x – 20 = 0 Solve by factoring
2. x2 + 3x - 4 = 0 Solve by completing the squares
3. 3x2 – x – 1 = 0 Solve by graphing (preferably using Excel)

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
I'll do the first one to get you started. If you're still stuck, then feel free to repost

# 1

First, let's factor the left side 3x%5E2%2B11x-20:


Looking at the expression 3x%5E2%2B11x-20, we can see that the first coefficient is 3, the second coefficient is 11, and the last term is -20.


Now multiply the first coefficient 3 by the last term -20 to get %283%29%28-20%29=-60.


Now the question is: what two whole numbers multiply to -60 (the previous product) and add to the second coefficient 11?


To find these two numbers, we need to list all of the factors of -60 (the previous product).


Factors of -60:
1,2,3,4,5,6,10,12,15,20,30,60
-1,-2,-3,-4,-5,-6,-10,-12,-15,-20,-30,-60


Note: list the negative of each factor. This will allow us to find all possible combinations.


These factors pair up and multiply to -60.
1*(-60)
2*(-30)
3*(-20)
4*(-15)
5*(-12)
6*(-10)
(-1)*(60)
(-2)*(30)
(-3)*(20)
(-4)*(15)
(-5)*(12)
(-6)*(10)

Now let's add up each pair of factors to see if one pair adds to the middle coefficient 11:


First NumberSecond NumberSum
1-601+(-60)=-59
2-302+(-30)=-28
3-203+(-20)=-17
4-154+(-15)=-11
5-125+(-12)=-7
6-106+(-10)=-4
-160-1+60=59
-230-2+30=28
-320-3+20=17
-415-4+15=11
-512-5+12=7
-610-6+10=4



From the table, we can see that the two numbers -4 and 15 add to 11 (the middle coefficient).


So the two numbers -4 and 15 both multiply to -60 and add to 11


Now replace the middle term 11x with -4x%2B15x. Remember, -4 and 15 add to 11. So this shows us that -4x%2B15x=11x.


3x%5E2%2Bhighlight%28-4x%2B15x%29-20 Replace the second term 11x with -4x%2B15x.


%283x%5E2-4x%29%2B%2815x-20%29 Group the terms into two pairs.


x%283x-4%29%2B%2815x-20%29 Factor out the GCF x from the first group.


x%283x-4%29%2B5%283x-4%29 Factor out 5 from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.


%28x%2B5%29%283x-4%29 Combine like terms. Or factor out the common term 3x-4


So 3x%5E2%2B11x-20 factors to %28x%2B5%29%283x-4%29.


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This means that 3x%5E2%2B11x-20=0 transforms into %28x%2B5%29%283x-4%29=0


%28x%2B5%29%283x-4%29=0 Start with the given equation.



Now set each factor equal to zero:

x%2B5=0 or 3x-4=0


Let's solve the first equation x%2B5=0


x=-5 Subtract 5 from both sides to isolate "x".

So the first solution is x=-5


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Now let's solve the second equation 3x-4=0


3x=4 Add 4 to both sides.


x=4%2F3 Divide both sides by 3 to isolate "x".


So the second solution is x=4%2F3


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Answer:


So the solutions are

x=-5 or x=4%2F3