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Mathematics 19 Online
OpenStudy (anonymous):

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)

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

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

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

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...

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?

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

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

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):

\(\huge~(4x^2-2\color{red}{x})+(10x+5)\)

OpenStudy (anonymous):

Are you trying to find the zeros?

OpenStudy (mathmate):

\(\huge~(4x^2-2\color{red}{x})\color{red}{-}(10x\color{red}{-}5)\)

pooja195 (pooja195):

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

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

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

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

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

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?

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

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

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?

pooja195 (pooja195):

please get this right -.-

OpenStudy (anonymous):

5/2 and 1/2

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