SOLUTION: solve for x: 25x^2 + 10x + 1 = 0

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Question 895924: solve for x: 25x^2 + 10x + 1 = 0
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%2B10x%2B1, we can see that the first coefficient is 25, the second coefficient is 10, and the last term is 1.



Now multiply the first coefficient 25 by the last term 1 to get %2825%29%281%29=25.



Now the question is: what two whole numbers multiply to 25 (the previous product) and add to the second coefficient 10?



To find these two numbers, we need to list all of the factors of 25 (the previous product).



Factors of 25:

1,5,25

-1,-5,-25



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



These factors pair up and multiply to 25.

1*25 = 25
5*5 = 25
(-1)*(-25) = 25
(-5)*(-5) = 25


Now let's add up each pair of factors to see if one pair adds to the middle coefficient 10:



First NumberSecond NumberSum
1251+25=26
555+5=10
-1-25-1+(-25)=-26
-5-5-5+(-5)=-10




From the table, we can see that the two numbers 5 and 5 add to 10 (the middle coefficient).



So the two numbers 5 and 5 both multiply to 25 and add to 10



Now replace the middle term 10x with 5x%2B5x. Remember, 5 and 5 add to 10. So this shows us that 5x%2B5x=10x.



25x%5E2%2Bhighlight%285x%2B5x%29%2B1 Replace the second term 10x with 5x%2B5x.



%2825x%5E2%2B5x%29%2B%285x%2B1%29 Group the terms into two pairs.



5x%285x%2B1%29%2B%285x%2B1%29 Factor out the GCF 5x from the first group.



5x%285x%2B1%29%2B1%285x%2B1%29 Factor out 1 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%2B1%29%285x%2B1%29 Combine like terms. Or factor out the common term 5x%2B1



%285x%2B1%29%5E2 Condense the terms.



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Answer:



So 25%2Ax%5E2%2B10%2Ax%2B1 factors to %285x%2B1%29%5E2.



In other words, 25%2Ax%5E2%2B10%2Ax%2B1=%285x%2B1%29%5E2.



Note: you can check the answer by expanding %285x%2B1%29%5E2 to get 25%2Ax%5E2%2B10%2Ax%2B1 or by graphing the original expression and the answer (the two graphs should be identical).