SOLUTION: Hey, Im in much need of help to figure out how to factor/solve qudaratic equations that are not in X^2+bx+c=0 form. for example,how do you factor 4x^2=4x+5?

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Question 449383: Hey, Im in much need of help to figure out how to factor/solve qudaratic equations that are not in X^2+bx+c=0 form. for example,how do you factor 4x^2=4x+5?
Found 2 solutions by MathLover1, stanbon:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!



factor 4x%5E2=4x%2B5..first write it in x%5E2%2Bbx%2Bc=0 form
4x%5E2-4x-5=0
Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)


Looking at the expression 4x%5E2-4x-5, we can see that the first coefficient is 4, the second coefficient is -4, and the last term is -5.



Now multiply the first coefficient 4 by the last term -5 to get %284%29%28-5%29=-20.



Now the question is: what two whole numbers multiply to -20 (the previous product) and add to the second coefficient -4?



To find these two numbers, we need to list all of the factors of -20 (the previous product).



Factors of -20:

1,2,4,5,10,20

-1,-2,-4,-5,-10,-20



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



These factors pair up and multiply to -20.

1*(-20) = -20
2*(-10) = -20
4*(-5) = -20
(-1)*(20) = -20
(-2)*(10) = -20
(-4)*(5) = -20


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



First NumberSecond NumberSum
1-201+(-20)=-19
2-102+(-10)=-8
4-54+(-5)=-1
-120-1+20=19
-210-2+10=8
-45-4+5=1




From the table, we can see that there are no pairs of numbers which add to -4. So 4x%5E2-4x-5 cannot be factored.



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Answer:



So 4%2Ax%5E2-4%2Ax-5 doesn't factor at all (over the rational numbers).



So 4%2Ax%5E2-4%2Ax-5 is prime.


Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
how do you factor 4x^2=4x+5?
---
Rearrange:
4x^2 - 4x - 5 = 0
----
Using the quadratic formula:
x = [4 +- sqrt(16-4*4*-5)]/(2*4)
---------------------
x = [4 +- sqrt(96)]/8
-----
x = [4 +- 4sqrt(6)]/8
----
r1 = (1/2)+(1/2)sqrt(6) or r2 = (1/2)-(1/2)sqrt(6)
-----
Each of these is a root of the quadratic.
---
The factored form would be (x-r1)(x-r2) = 0
===============================================
So the quadratic formula can ALWAYS be used to factor
a quadratic polynomial.
===============================================
If you are familiar with FOIL, Inverse FOIL can
be used to factor some quadratics.
----
Example:
2x^2+7x-15 = 0
ac = 2*-15 = -30
b = 7
----
Think of 2 numbers whose product is ac and whose sum is b:
ac = 10*-3 = -30
b = 10+ -3 = 7
-----
Rewrite the quadratic replacing the "b-term":
2x^2 + 10x-3x -15 = 0
-----
Factor the 1st two and the last two terms separately:
2x(x+5)-3(x+5) = 0
Factor again:
(x+5)(2x-3) = 0
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Cheers,
Stan H.
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