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

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

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

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ik the forumla is this: x = [ -b ± sqrt(b^2 - 4ac) ] / 2a
well we should use complete the square method you can move the constant term to the right side
x is the term correct?
\[ \huge\rm 4x^2 − 12x + 3 = 0\] subtract 3 both sides
not just x term are divided by plus or negative sign
uh... subtract the 4x^2 by 3 and the 12 by 3?
\[\huge\rm Ax^2+Bx+C=0\]quadratic e quation where A=leading coefficient B=middle term C=constant term(number with no variable )
|dw:1443453772906:dw| you can't combine -3 and 4x^2 they arn't like terms
im not sure what you do next after you get -3..
divide?
|dw:1443453908051:dw| we should complete the square of the left part
\[\huge\rm \color{Red}{4x^2+12x}=-3\] what is the common factor at left side ?
the x?
no just take out the common number not the varaible
uh ... im confused. am i supposed to take out the 4 and 12 and divide them?
besides the x i see nothing in common...
well it's 4x^2+12x so we just need to take out the common number
what are the factors of 4 and 12 ?
1, 2, 4 ... 4 is best factor... then i divide both by four to get 3 right?
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 ?
greatest number **
uh i said 4...
is there one greater than thta?
that*
i don't understand what u mean "divide `both` to get 3"
\[\huge\rm 4(x^2+3x)=-3\]when you take out 4 from 4 u will get 1
you still have the 12x
i was saying didvide the 4x^2 and 12x by the common factor to simplify the solution.
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
yes right. i read it wrong. so sorry
\[\huge\rm 4(x^2+3x)=-3\] take half of the x term and then square it add square to the both sides
you have to x's which one do i take half of?
two*
first one is x^2 2nd one is x
so take half of the second one?
yes right
uh... ok so now you'll have- x+ 3?
thats not a negative sign
x+3 how ? we should take half of 3 in other words just divide `b by 2`
how am i dividing 3x by 2?
/i thought you wanted me to take away x on both sides..
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
well first we need to divide b by 2 take square of (b/2)^2 then add (b/2)^2 to both sides
what is b in ur equation ?
the 3x?
just b
3 =b not 3x x is variable
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
do we get rid of negative 3 with the positive 3
positive 3 ?? it's 3x you can't combine 3x with -3 they are not like terms
then your left with whats in blue right?
right we can factor this quadraticc equation (x^2 +3/2x + (3/2)^2) or look at this example \[\huge\rm (x^2+bx) =(x+\frac{ b }{ 2 })^2\]
you need to convert left side in squared form
so its a bit of a leap here but i believe the simplified form would be -3 +- 2\[\sqrt{6}\] over 2?
everything over 2*
and how did you get that ?
well you told me that like terms cant combine
leaving negative 3 alone.
look at this example \[\huge\rm (x^2+bx) =(x+\frac{ b }{ 2 })^2\] and tell me how would you write \[(x^2+3x) = ????\]
then you said something about b being divided by2? which i believe gave me that..
how did you write \[(x^2+3x)\] in squared form ?
mmm
3x^2? idk
ik your left with one x right?
\(\color{blue}{\text{Originally Posted by}}\) @Nnesha look at this example \[\huge\rm (x^2+bx) =(x+\frac{ b }{ 2 })^2\] and tell me how would you write \[(x^2+3x) = ????\] \(\color{blue}{\text{End of Quote}}\)
thats where im lost though...
yea that's obvious we have to solve for x
then how you got the final x value ?
i got what?
here is another example \[(x^2+6x)=(x+\frac{ 6 }{ 2})^2=(x+3)^2\]
wait but isn't that similar to what i had?
similar to what ?
\[\huge\rm 4(x^2+\color{blue}{\frac{ 3x }{ 2 }}+\color{Red}{(\frac{ 3 }{ 2 })^2}) = -3 +\color{red}{(\frac{3}{2})^2 } \times 4\] \[4(x+\frac{ 3 }{ 2})^2=-3+(\frac{ 3 }{ 2 })^2 \times 4\] according to what i said (x^2+bx)=(x+b/2)^2
solve this to get the answer \[4(x+\frac{ 3 }{ 2})^2=-3+(\frac{ 3 }{ 2 })^2 \times 4\]
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 \[(x^2+bx)=(x+\frac{ b }{ 2})^2\]\[(x^2+6x)=(x+\frac{ 6 }{ 2 })^2=(x+3)^2\]
feel free to ask question about that^let me know if you have question on that
ok thank you
i had another one mind helping?
did you get this one ?
i believe i did
Solve x2 − 5x + 1 = 0 using the quadratic formula.
alright i'll try
i know the first step to this ...
can u make a new post bec this one is laggy
i have noticed
ye it keeps moving hard to use latex on this post

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