SOLUTION: y=25x^2+55x+24

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Question 280744: y=25x^2+55x+24
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)


Looking at the expression 25x%5E2%2B55x%2B24, we can see that the first coefficient is 25, the second coefficient is 55, and the last term is 24.



Now multiply the first coefficient 25 by the last term 24 to get %2825%29%2824%29=600.



Now the question is: what two whole numbers multiply to 600 (the previous product) and add to the second coefficient 55?



To find these two numbers, we need to list all of the factors of 600 (the previous product).



Factors of 600:

1,2,3,4,5,6,8,10,12,15,20,24,25,30,40,50,60,75,100,120,150,200,300,600

-1,-2,-3,-4,-5,-6,-8,-10,-12,-15,-20,-24,-25,-30,-40,-50,-60,-75,-100,-120,-150,-200,-300,-600



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



These factors pair up and multiply to 600.

1*600 = 600
2*300 = 600
3*200 = 600
4*150 = 600
5*120 = 600
6*100 = 600
8*75 = 600
10*60 = 600
12*50 = 600
15*40 = 600
20*30 = 600
24*25 = 600
(-1)*(-600) = 600
(-2)*(-300) = 600
(-3)*(-200) = 600
(-4)*(-150) = 600
(-5)*(-120) = 600
(-6)*(-100) = 600
(-8)*(-75) = 600
(-10)*(-60) = 600
(-12)*(-50) = 600
(-15)*(-40) = 600
(-20)*(-30) = 600
(-24)*(-25) = 600


Now let's add up each pair of factors to see if one pair adds to the middle coefficient 55:



First NumberSecond NumberSum
16001+600=601
23002+300=302
32003+200=203
41504+150=154
51205+120=125
61006+100=106
8758+75=83
106010+60=70
125012+50=62
154015+40=55
203020+30=50
242524+25=49
-1-600-1+(-600)=-601
-2-300-2+(-300)=-302
-3-200-3+(-200)=-203
-4-150-4+(-150)=-154
-5-120-5+(-120)=-125
-6-100-6+(-100)=-106
-8-75-8+(-75)=-83
-10-60-10+(-60)=-70
-12-50-12+(-50)=-62
-15-40-15+(-40)=-55
-20-30-20+(-30)=-50
-24-25-24+(-25)=-49




From the table, we can see that the two numbers 15 and 40 add to 55 (the middle coefficient).



So the two numbers 15 and 40 both multiply to 600 and add to 55



Now replace the middle term 55x with 15x%2B40x. Remember, 15 and 40 add to 55. So this shows us that 15x%2B40x=55x.



25x%5E2%2Bhighlight%2815x%2B40x%29%2B24 Replace the second term 55x with 15x%2B40x.



%2825x%5E2%2B15x%29%2B%2840x%2B24%29 Group the terms into two pairs.



5x%285x%2B3%29%2B%2840x%2B24%29 Factor out the GCF 5x from the first group.



5x%285x%2B3%29%2B8%285x%2B3%29 Factor out 8 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.



%285x%2B8%29%285x%2B3%29 Combine like terms. Or factor out the common term 5x%2B3



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Answer:



So 25%2Ax%5E2%2B55%2Ax%2B24 factors to %285x%2B8%29%285x%2B3%29.



In other words, 25%2Ax%5E2%2B55%2Ax%2B24=%285x%2B8%29%285x%2B3%29.



Note: you can check the answer by expanding %285x%2B8%29%285x%2B3%29 to get 25%2Ax%5E2%2B55%2Ax%2B24 or by graphing the original expression and the answer (the two graphs should be identical).


Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 25x%5E2%2B55x%2B24+=+0) has the following solutons:

x%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=%2855%29%5E2-4%2A25%2A24=625.

Discriminant d=625 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-55%2B-sqrt%28+625+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%2855%29%2Bsqrt%28+625+%29%29%2F2%5C25+=+-0.6
x%5B2%5D+=+%28-%2855%29-sqrt%28+625+%29%29%2F2%5C25+=+-1.6

Quadratic expression 25x%5E2%2B55x%2B24 can be factored:
25x%5E2%2B55x%2B24+=+25%28x--0.6%29%2A%28x--1.6%29
Again, the answer is: -0.6, -1.6. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+25%2Ax%5E2%2B55%2Ax%2B24+%29