Solve the equation 4x2 − 12x + 3 = 0 by completing the square. Verify your answer using the quadratic formula.

- some.random.cool.kid

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- some.random.cool.kid

@Nnesha

- some.random.cool.kid

@nincompoop

- some.random.cool.kid

i have a few of these this is just the first that i need help with...

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## More answers

- some.random.cool.kid

@ParthKohli

- some.random.cool.kid

@peachpi

- some.random.cool.kid

@whpalmer4

- some.random.cool.kid

ik the forumla is this:
x = [ -b ± sqrt(b^2 - 4ac) ] / 2a

- Nnesha

well we should use complete the square method
you can move the constant term to the right side

- some.random.cool.kid

x is the term correct?

- Nnesha

\[ \huge\rm 4x^2 − 12x + 3 = 0\]
subtract 3 both sides

- Nnesha

not just x
term are divided by plus or negative sign

- some.random.cool.kid

uh... subtract the 4x^2 by 3 and the 12 by 3?

- Nnesha

\[\huge\rm Ax^2+Bx+C=0\]quadratic e quation where
A=leading coefficient
B=middle term
C=constant term(number with no variable )

- Nnesha

|dw:1443453772906:dw|
you can't combine -3 and 4x^2 they arn't like terms

- some.random.cool.kid

im not sure what you do next after you get -3..

- some.random.cool.kid

divide?

- Nnesha

|dw:1443453908051:dw|
we should complete the square of the left part

- Nnesha

\[\huge\rm \color{Red}{4x^2+12x}=-3\]
what is the common factor at left side ?

- some.random.cool.kid

the x?

- Nnesha

no just take out the common number not the varaible

- some.random.cool.kid

uh ... im confused. am i supposed to take out the 4 and 12 and divide them?

- some.random.cool.kid

besides the x i see nothing in common...

- Nnesha

well it's 4x^2+12x
so we just need to take out the common number

- Nnesha

what are the factors of 4 and 12 ?

- some.random.cool.kid

1, 2, 4 ... 4 is best factor... then i divide both by four to get 3 right?

- Nnesha

you need GREATEST COMMON FACTOR
1,2,4 are factors of 4
1,2,3,4,6,12 are factors of 12
so which number is common ?

- Nnesha

greatest number **

- some.random.cool.kid

uh i said 4...

- some.random.cool.kid

is there one greater than thta?

- some.random.cool.kid

that*

- Nnesha

i don't understand what u mean "divide `both` to get 3"

- Nnesha

\[\huge\rm 4(x^2+3x)=-3\]when you take out 4 from 4 u will get 1

- some.random.cool.kid

you still have the 12x

- some.random.cool.kid

i was saying didvide the 4x^2 and 12x by the common factor to simplify the solution.

- Nnesha

now we can complete the square of (x^2+3x)
we should divide b by b
here is an example \[\huge\rm a(x^2+bx)=-C\]
divide b by 2 \[\huge\rm a(x+\frac{ b }{ 2 })^2=-C + (\frac{ b }{ 2})^2 \]
and then add(b/2)^2 to both sides

- Nnesha

yes right.
i read it wrong. so sorry

- Nnesha

\[\huge\rm 4(x^2+3x)=-3\]
take half of the x term and then square it
add square to the both sides

- some.random.cool.kid

you have to x's which one do i take half of?

- some.random.cool.kid

two*

- Nnesha

first one is x^2
2nd one is x

- some.random.cool.kid

so take half of the second one?

- Nnesha

yes right

- some.random.cool.kid

uh... ok so now you'll have- x+ 3?

- some.random.cool.kid

thats not a negative sign

- Nnesha

x+3 how ?
we should take half of 3
in other words just divide `b by 2`

- some.random.cool.kid

how am i dividing 3x by 2?

- some.random.cool.kid

/i thought you wanted me to take away x on both sides..

- Nnesha

here is an example \[\huge\rm a(x^2+bx)=-C\]
divide b by 2 \[\huge\rm a(x+\frac{ b }{ 2 })^2=-C + (\frac{ b }{ 2})^2 \times a \]
and then add(b/2)^2 to both sides
here is an example

- Nnesha

well first we need to divide b by 2
take square of (b/2)^2
then add (b/2)^2 to both sides

- Nnesha

what is b in ur equation ?

- some.random.cool.kid

the 3x?

- Nnesha

just b

- Nnesha

3 =b
not 3x x is variable

- Nnesha

yes right \[\huge\rm 4(x^2+\color{blue}{\frac{ 3x }{ 2 }}+\color{Red}{(\frac{ 3 }{ 2 })^2}) = -3 +\color{red}{(\frac{3}{2})^2 } \times 4\]
look at the blue term i took half of 3
and then add (b/2)^2 to both sides
and sicne we got the common factor we should multiply (b/2)^2 by 4 at right side

- some.random.cool.kid

do we get rid of negative 3 with the positive 3

- Nnesha

positive 3 ?? it's 3x you can't combine 3x with -3 they are not like terms

- some.random.cool.kid

then your left with whats in blue right?

- Nnesha

right we can factor this quadraticc equation (x^2 +3/2x + (3/2)^2)
or look at this example \[\huge\rm (x^2+bx) =(x+\frac{ b }{ 2 })^2\]

- Nnesha

you need to convert left side in squared form

- some.random.cool.kid

so its a bit of a leap here but i believe the simplified form would be -3 +- 2\[\sqrt{6}\] over 2?

- some.random.cool.kid

everything over 2*

- Nnesha

and how did you get that ?

- some.random.cool.kid

well you told me that like terms cant combine

- some.random.cool.kid

leaving negative 3 alone.

- Nnesha

look at this example \[\huge\rm (x^2+bx) =(x+\frac{ b }{ 2 })^2\]
and tell me how would you write \[(x^2+3x) = ????\]

- some.random.cool.kid

then you said something about b being divided by2? which i believe gave me that..

- Nnesha

how did you write \[(x^2+3x)\] in squared form ?

- some.random.cool.kid

mmm

- some.random.cool.kid

3x^2? idk

- some.random.cool.kid

ik your left with one x right?

- Nnesha

\(\color{blue}{\text{Originally Posted by}}\) @Nnesha
look at this example \[\huge\rm (x^2+bx) =(x+\frac{ b }{ 2 })^2\]
and tell me how would you write \[(x^2+3x) = ????\]
\(\color{blue}{\text{End of Quote}}\)

- some.random.cool.kid

thats where im lost though...

- Nnesha

yea that's obvious we have to solve for x

- Nnesha

then how you got the final x value ?

- some.random.cool.kid

i got what?

- Nnesha

here is another example \[(x^2+6x)=(x+\frac{ 6 }{ 2})^2=(x+3)^2\]

- some.random.cool.kid

wait but isn't that similar to what i had?

- Nnesha

similar to what ?

- Nnesha

\[\huge\rm 4(x^2+\color{blue}{\frac{ 3x }{ 2 }}+\color{Red}{(\frac{ 3 }{ 2 })^2}) = -3 +\color{red}{(\frac{3}{2})^2 } \times 4\]
\[4(x+\frac{ 3 }{ 2})^2=-3+(\frac{ 3 }{ 2 })^2 \times 4\]
according to what i said
(x^2+bx)=(x+b/2)^2

- Nnesha

solve this to get the answer \[4(x+\frac{ 3 }{ 2})^2=-3+(\frac{ 3 }{ 2 })^2 \times 4\]

- Nnesha

here is how we can write in squared form \[(x^2+bx)=(x+\frac{ b }{ 2 })^2\]
\[(x^2+6x)=x^2+6x+\color{reD}{9}\]
i divided 6 by 2 which is equal to 3 then squared 3
that's how i got 9
now we can factor x^2+6x+9
(x+3)(x+3) are factors of that quadratic equation
which is same as (x+3)^2
so in short we can do this \[(x^2+bx)=(x+\frac{ b }{ 2})^2\]\[(x^2+6x)=(x+\frac{ 6 }{ 2 })^2=(x+3)^2\]

- Nnesha

feel free to ask question about that^let me know if you have question on that

- some.random.cool.kid

ok thank you

- some.random.cool.kid

i had another one mind helping?

- some.random.cool.kid

@Nnesha

- Nnesha

did you get this one ?

- some.random.cool.kid

i believe i did

- some.random.cool.kid

Solve x2 − 5x + 1 = 0 using the quadratic formula.

- Nnesha

alright i'll try

- some.random.cool.kid

i know the first step to this ...

- Nnesha

can u make a new post bec this one is laggy

- some.random.cool.kid

i have noticed

- Nnesha

ye it keeps moving
hard to use latex on this post

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