PLEASE HELP I WILL GIVE MEDAL
Part B: Solve 4x2 -12x + 5 = 0 using an appropriate method. Show the steps of your work, and explain why you chose the method used. (4 points)
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pooja195 (pooja195):
Have you learned factoring?
OpenStudy (anonymous):
This is an example of factoring trinomials :)
OpenStudy (anonymous):
can you please help
OpenStudy (anonymous):
im stuck and need to be done within the hour
OpenStudy (anonymous):
this one factors
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OpenStudy (anonymous):
ok
OpenStudy (anonymous):
do you know how it factors?
OpenStudy (anonymous):
ehh not really :(
OpenStudy (anonymous):
me neither
i cheat
OpenStudy (anonymous):
:( lol
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pooja195 (pooja195):
|dw:1433297189737:dw|
OpenStudy (anonymous):
factors as
\[(2x-1)(2x-5)=0\] would you like to know how i know it?
pooja195 (pooja195):
Or that...
OpenStudy (anonymous):
yes please!
OpenStudy (anonymous):
i will wait and let @pooja195 explain
maybe i will learn something
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OpenStudy (anonymous):
i have never figured out a way to factor other than to grind it til i find it
pooja195 (pooja195):
thanks :)
Ok well fist you want to multiply the a and c value what do you get?
OpenStudy (anonymous):
this is algebra 1 incase you where wondering!
pooja195 (pooja195):
Whats
\[\huge 5 \times 4 \]
OpenStudy (anonymous):
20...
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pooja195 (pooja195):
good and our B value is -12 right?
OpenStudy (anonymous):
yes sir
pooja195 (pooja195):
|dw:1433297395133:dw|
OpenStudy (anonymous):
ok
pooja195 (pooja195):
Now ask yourself what multiplies into -20 BUT also adds up to -12?
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pooja195 (pooja195):
You will have 2 numbers...
OpenStudy (anonymous):
2 and 10?
OpenStudy (mathmate):
multiplies into 20...
OpenStudy (anonymous):
or 4 and 5
pooja195 (pooja195):
oh right multiplies into 20 but adds to -12
you are close with 10 and 2
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pooja195 (pooja195):
you need a sign on both numbers...
OpenStudy (anonymous):
-2 and -10?
OpenStudy (anonymous):
or -2 and 10
pooja195 (pooja195):
yes! :)
Now...
\[\huge~(4x^2-2)+(10x+5)\]
pooja195 (pooja195):
Now find the GCF
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pooja195 (pooja195):
for each...
OpenStudy (anonymous):
After @pooja195 is done, I can summarize it all into one huge message of steps so it won't seem all over the place :)
OpenStudy (anonymous):
-2 for the first and 5 for the second?
OpenStudy (anonymous):
thank you aureyliant!
OpenStudy (mathmate):
\(\huge~(4x^2-2\color{red}{x})+(10x+5)\)
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You have the signs messed up but 2x and 5 is right :)
\[\huge 2x(2x-1)+5(2x-1)\]
pooja195 (pooja195):
*minus
pooja195 (pooja195):
xD
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OpenStudy (anonymous):
so 2x(2x-1) - 5 (2x-1)?
OpenStudy (anonymous):
i know but im making sure that was correct XD
OpenStudy (anonymous):
(2x - 5)(2x - 1)
pooja195 (pooja195):
now you just combine the parenthese and the outter numbers
(2x-5)(2x-1)
and you end up with Sat's grinding :P
OpenStudy (anonymous):
You need to combine the first loner and the second loner and bring them together
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OpenStudy (anonymous):
ok is that all to it?
pooja195 (pooja195):
Then you set both equations to 0
OpenStudy (anonymous):
Trying to factor by splitting the middle term
2.1 Factoring 4x2-12x+5
The first term is, 4x2 its coefficient is 4 .
The middle term is, -12x its coefficient is -12 .
The last term, "the constant", is +5
Step-1 : Multiply the coefficient of the first term by the constant 4 • 5 = 20
Step-2 : Find two factors of 20 whose sum equals the coefficient of the middle term, which is -12 .
-20 + -1 = -21
-10 + -2 = -12 That's it
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -10 and -2
4x2 - 10x - 2x - 5
Step-4 : Add up the first 2 terms, pulling out like factors :
2x • (2x-5)
Add up the last 2 terms, pulling out common factors :
1 • (2x-5)
Step-5 : Add up the four terms of step 4 :
(2x-1) • (2x-5)
Which is the desired factorization
Equation at the end of step 2 :
(2x - 5) • (2x - 1) = 0
Step 3 :
Solve (2x-5)•(2x-1) = 0
(This goes towards solving for the zeros but that's only if the Q is asking that, too)
pooja195 (pooja195):
2x-5=0
2x-1=0
OpenStudy (anonymous):
^^ Yup
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OpenStudy (anonymous):
ok
pooja195 (pooja195):
Solve it .-.
OpenStudy (anonymous):
Now you could either do that or use the quadratic formula
OpenStudy (anonymous):
But thats a story for another day
pooja195 (pooja195):
^no
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pooja195 (pooja195):
Just set it to 0 its sooooo much more easier
OpenStudy (anonymous):
Solving a Single Variable Equation :
3.2 Solve : 2x-5 = 0
Add 5 to both sides of the equation :
2x = 5
Divide both sides of the equation by 2:
x = 5/2 = 2.500
Solving a Single Variable Equation :
3.3 Solve : 2x-1 = 0
Add 1 to both sides of the equation :
2x = 1
Divide both sides of the equation by 2:
x = 1/2 = 0.500
OpenStudy (anonymous):
can you guide me through setting through 0?
OpenStudy (anonymous):
It depends on the type of factoring Q Pooja
pooja195 (pooja195):
2x-5=0
we want to isolate X
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OpenStudy (anonymous):
It may not be as easy to split the middle term
OpenStudy (anonymous):
ok
pooja195 (pooja195):
so start by adding 5
pooja195 (pooja195):
that way on the left side the 5 will cancel what does the equation look like now?
OpenStudy (anonymous):
ok so 2x=5?
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pooja195 (pooja195):
Yes now just divde both sides by 2
pooja195 (pooja195):
what do you get?
OpenStudy (anonymous):
1x=1/2????
pooja195 (pooja195):
No
pooja195 (pooja195):
keep the 5
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OpenStudy (anonymous):
Now we know the two values for x... also I summarized solving for the zeros above @cacaman
pooja195 (pooja195):
you get x=5/2
OpenStudy (anonymous):
X=5?
OpenStudy (anonymous):
just kidding
pooja195 (pooja195):
-_- i gave u the answer :P
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pooja195 (pooja195):
2x-1=0
solve
pooja195 (pooja195):
I think you answered this one earlier...
OpenStudy (anonymous):
2x=1
OpenStudy (anonymous):
so 1/2
pooja195 (pooja195):
YES!
now what are the 2 answers?
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