SOLUTION: What are the different methods for solving quadratic equations? Give an example. Please help me and thank you!

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: What are the different methods for solving quadratic equations? Give an example. Please help me and thank you!      Log On


   

Question 897456: What are the different methods for solving quadratic equations? Give an example. Please help me and thank you!
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
Factor [Factorize]
Complete the square
Use quadratic formula

Factor by grouping
Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)


2%2Ax%5E2%2B8%2Ax%2B6 Start with the given expression.



2%28x%5E2%2B4x%2B3%29 Factor out the GCF 2.



Now let's try to factor the inner expression x%5E2%2B4x%2B3



---------------------------------------------------------------



Looking at the expression x%5E2%2B4x%2B3, we can see that the first coefficient is 1, the second coefficient is 4, and the last term is 3.



Now multiply the first coefficient 1 by the last term 3 to get %281%29%283%29=3.



Now the question is: what two whole numbers multiply to 3 (the previous product) and add to the second coefficient 4?



To find these two numbers, we need to list all of the factors of 3 (the previous product).



Factors of 3:

1,3

-1,-3



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



These factors pair up and multiply to 3.

1*3 = 3
(-1)*(-3) = 3


Now let's add up each pair of factors to see if one pair adds to the middle coefficient 4:



First NumberSecond NumberSum
131+3=4
-1-3-1+(-3)=-4




From the table, we can see that the two numbers 1 and 3 add to 4 (the middle coefficient).



So the two numbers 1 and 3 both multiply to 3 and add to 4



Now replace the middle term 4x with x%2B3x. Remember, 1 and 3 add to 4. So this shows us that x%2B3x=4x.



x%5E2%2Bhighlight%28x%2B3x%29%2B3 Replace the second term 4x with x%2B3x.



%28x%5E2%2Bx%29%2B%283x%2B3%29 Group the terms into two pairs.



x%28x%2B1%29%2B%283x%2B3%29 Factor out the GCF x from the first group.



x%28x%2B1%29%2B3%28x%2B1%29 Factor out 3 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%2B3%29%28x%2B1%29 Combine like terms. Or factor out the common term x%2B1



--------------------------------------------------



So 2%28x%5E2%2B4x%2B3%29 then factors further to 2%28x%2B3%29%28x%2B1%29



===============================================================



Answer:



So 2%2Ax%5E2%2B8%2Ax%2B6 completely factors to 2%28x%2B3%29%28x%2B1%29.



In other words, 2%2Ax%5E2%2B8%2Ax%2B6=2%28x%2B3%29%28x%2B1%29.



Note: you can check the answer by expanding 2%28x%2B3%29%28x%2B1%29 to get 2%2Ax%5E2%2B8%2Ax%2B6 or by graphing the original expression and the answer (the two graphs should be identical).


Complete the square
Solved by pluggable solver: COMPLETING THE SQUARE solver for quadratics
Read this lesson on completing the square by prince_abubu, if you do not know how to complete the square.
Let's convert 1x%5E2%2B-1x%2B-2=0 to standard form by dividing both sides by 1:
We have: 1x%5E2%2B-1x%2B-2=0. What we want to do now is to change this equation to a complete square %28x%2Bsomenumber%29%5E2+%2B+othernumber. How can we find out values of somenumber and othernumber that would make it work?
Look at %28x%2Bsomenumber%29%5E2: %28x%2Bsomenumber%29%5E2+=+x%5E2%2B2%2Asomenumber%2Ax+%2B+somenumber%5E2. Since the coefficient in our equation 1x%5E2%2Bhighlight_red%28+-1%29+%2A+x%2B-2=0 that goes in front of x is -1, we know that -1=2*somenumber, or somenumber+=+-1%2F2. So, we know that our equation can be rewritten as %28x%2B-1%2F2%29%5E2+%2B+othernumber, and we do not yet know the other number.
We are almost there. Finding the other number is simply a matter of not making too many mistakes. We need to find 'other number' such that %28x%2B-1%2F2%29%5E2+%2B+othernumber is equivalent to our original equation 1x%5E2%2B-1x%2Bhighlight_green%28+-2+%29=0.


The highlighted red part must be equal to -2 (highlighted green part).

-1%5E2%2F4+%2B+othernumber+=+-2, or othernumber+=+-2--1%5E2%2F4+=+-2.25.
So, the equation converts to %28x%2B-1%2F2%29%5E2+%2B+-2.25+=+0, or %28x%2B-1%2F2%29%5E2+=+2.25.

Our equation converted to a square %28x%2B-1%2F2%29%5E2, equated to a number (2.25).

Since the right part 2.25 is greater than zero, there are two solutions:

system%28+%28x%2B-1%2F2%29+=+%2Bsqrt%28+2.25+%29%2C+%28x%2B-1%2F2%29+=+-sqrt%28+2.25+%29+%29
, or

system%28+%28x%2B-1%2F2%29+=+1.5%2C+%28x%2B-1%2F2%29+=+-1.5+%29
system%28+x%2B-1%2F2+=+1.5%2C+x%2B-1%2F2+=+-1.5+%29
system%28+x+=+1.5--1%2F2%2C+x+=+-1.5--1%2F2+%29

system%28+x+=+2%2C+x+=+-1+%29
Answer: x=2, -1.

Use quadratic formula
Solved by pluggable solver: Quadratic Formula
Let's use the quadratic formula to solve for x:


Starting with the general quadratic


ax%5E2%2Bbx%2Bc=0


the general solution using the quadratic equation is:


x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29




So lets solve x%5E2%2B2%2Ax%2B3=0 ( notice a=1, b=2, and c=3)





x+=+%28-2+%2B-+sqrt%28+%282%29%5E2-4%2A1%2A3+%29%29%2F%282%2A1%29 Plug in a=1, b=2, and c=3




x+=+%28-2+%2B-+sqrt%28+4-4%2A1%2A3+%29%29%2F%282%2A1%29 Square 2 to get 4




x+=+%28-2+%2B-+sqrt%28+4%2B-12+%29%29%2F%282%2A1%29 Multiply -4%2A3%2A1 to get -12




x+=+%28-2+%2B-+sqrt%28+-8+%29%29%2F%282%2A1%29 Combine like terms in the radicand (everything under the square root)




x+=+%28-2+%2B-+2%2Ai%2Asqrt%282%29%29%2F%282%2A1%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)




x+=+%28-2+%2B-+2%2Ai%2Asqrt%282%29%29%2F%282%29 Multiply 2 and 1 to get 2




After simplifying, the quadratic has roots of


x=-1%2Bsqrt%282%29i or x=-1-sqrt%282%29i