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)

- anonymous

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- pooja195

Have you learned factoring?

- anonymous

This is an example of factoring trinomials :)

- anonymous

can you please help

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

im stuck and need to be done within the hour

- anonymous

this one factors

- anonymous

ok

- anonymous

do you know how it factors?

- anonymous

ehh not really :(

- anonymous

me neither
i cheat

- anonymous

:( lol

- pooja195

|dw:1433297189737:dw|

- anonymous

factors as
\[(2x-1)(2x-5)=0\] would you like to know how i know it?

- pooja195

Or that...

- anonymous

yes please!

- anonymous

i will wait and let @pooja195 explain
maybe i will learn something

- anonymous

i have never figured out a way to factor other than to grind it til i find it

- pooja195

thanks :)
Ok well fist you want to multiply the a and c value what do you get?

- anonymous

this is algebra 1 incase you where wondering!

- pooja195

Whats
\[\huge 5 \times 4 \]

- anonymous

20...

- pooja195

good and our B value is -12 right?

- anonymous

yes sir

- pooja195

|dw:1433297395133:dw|

- anonymous

ok

- pooja195

Now ask yourself what multiplies into -20 BUT also adds up to -12?

- pooja195

You will have 2 numbers...

- anonymous

2 and 10?

- mathmate

multiplies into 20...

- anonymous

or 4 and 5

- pooja195

oh right multiplies into 20 but adds to -12
you are close with 10 and 2

- pooja195

you need a sign on both numbers...

- anonymous

-2 and -10?

- anonymous

or -2 and 10

- pooja195

yes! :)
Now...
\[\huge~(4x^2-2)+(10x+5)\]

- pooja195

Now find the GCF

- pooja195

for each...

- 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 :)

- anonymous

-2 for the first and 5 for the second?

- anonymous

thank you aureyliant!

- mathmate

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

- anonymous

Are you trying to find the zeros?

- mathmate

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

- pooja195

You have the signs messed up but 2x and 5 is right :)
\[\huge 2x(2x-1)+5(2x-1)\]

- pooja195

*minus

- pooja195

xD

- anonymous

so 2x(2x-1) - 5 (2x-1)?

- anonymous

i know but im making sure that was correct XD

- anonymous

(2x - 5)(2x - 1)

- pooja195

now you just combine the parenthese and the outter numbers
(2x-5)(2x-1)
and you end up with Sat's grinding :P

- anonymous

You need to combine the first loner and the second loner and bring them together

- anonymous

ok is that all to it?

- pooja195

Then you set both equations to 0

- 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 â€¢ 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 â€¢ (2x-5)
Add up the last 2 terms, pulling out common factors :
1 â€¢ (2x-5)
Step-5 : Add up the four terms of step 4 :
(2x-1) â€¢ (2x-5)
Which is the desired factorization
Equation at the end of step 2 :
(2x - 5) â€¢ (2x - 1) = 0
Step 3 :
Solve (2x-5)â€¢(2x-1) = 0
(This goes towards solving for the zeros but that's only if the Q is asking that, too)

- pooja195

2x-5=0
2x-1=0

- anonymous

^^ Yup

- anonymous

ok

- pooja195

Solve it .-.

- anonymous

Now you could either do that or use the quadratic formula

- anonymous

But thats a story for another day

- pooja195

^no

- pooja195

Just set it to 0 its sooooo much more easier

- 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

- anonymous

can you guide me through setting through 0?

- anonymous

It depends on the type of factoring Q Pooja

- pooja195

2x-5=0
we want to isolate X

- anonymous

It may not be as easy to split the middle term

- anonymous

ok

- pooja195

so start by adding 5

- pooja195

that way on the left side the 5 will cancel what does the equation look like now?

- anonymous

ok so 2x=5?

- pooja195

Yes now just divde both sides by 2

- pooja195

what do you get?

- anonymous

1x=1/2????

- pooja195

No

- pooja195

keep the 5

- anonymous

Now we know the two values for x... also I summarized solving for the zeros above @cacaman

- pooja195

you get x=5/2

- anonymous

X=5?

- anonymous

just kidding

- pooja195

-_- i gave u the answer :P

- pooja195

2x-1=0
solve

- pooja195

I think you answered this one earlier...

- anonymous

2x=1

- anonymous

so 1/2

- pooja195

YES!
now what are the 2 answers?

- pooja195

please get this right -.-

- anonymous

5/2 and 1/2

- pooja195

Good! :)

- anonymous

lol im nervous!

- pooja195

why so? :P
and why did we choose this do you know?

- anonymous

no..

- pooja195

You tell me because i dont know either.

- anonymous

LOL

- anonymous

We chose this because it was an easier method in my books?

- pooja195

Find a better reason -_-

- pooja195

I guess you can write because it is a easier way to do it :/ and since its a trinomial grouping is involved...

- Nnesha

there is another method to factor quadratic equation
cross multiply instead group method |dw:1433299203705:dw|
factor first and last term
then cross multiply after that combine them if you get middle term
then (2x-5) and (2x-1) are factor of that equation

Looking for something else?

Not the answer you are looking for? Search for more explanations.