Question

3x^2-9x-12=0 by completing the square
with explanation if possible.

Answer 1

divide the whole thing by 3

Answer 2

x^2-3x-4=0

Answer 3

\[3(x^2-3x-4) = 0 \rightarrow 3(x^2-3x)=4\]

Answer 4

to complete the square...you take the middle term and divide by 2 and square it

Answer 5

\[3(x^2-3x+(\frac{ 3 }{ 2 })^2)=4+(\frac{ 3 }{ 2 })^2\]

Answer 6

looks like i forgot to x3 on the other end

Answer 7

\[3(x^2-3x+(\frac{ 3 }{ 2 })^2=12+3(\frac{ 3 }{ 2 })^2\]

Answer 8

its 3 not 4

Answer 9

and also completing the square method

Answer 10

you should get \[3(x^2-3x+(\frac{ 9 }{ 4 })=18\frac{ 3 }{ 4 }\]

Answer 11

now you can factorize the\[3(x^2-3x+\frac{ 9 }{ 4 })\]part

Answer 12

you catching on?

Answer 13

yes, thank you very much

Answer 14

you can factorize that on your own?

Answer 15

yes, i was having trouble getting to that point though. appreciate your help!

Answer 16

yw

Answer 17

Start with your equation:

\(3x^2-9x-12=0 \)

Notice that there is a common factor of 3 for all terms, so divide both sides by 3:

\(x^2-3x-4=0 \)

Now add 4 to both sides.

\(x^2-3x=4 \)

The term that needs to be added to the left sides is obtained this way. Take half of the x-term coefficient and square it.
Half of -3 is \(-\dfrac{3}{2}\). The square of that is \(\dfrac{9}{4} \)

Now we add \(\dfrac{9}{4} \) to both sides

\(x^2-3x + \dfrac{9}{4}=4 + \dfrac{9}{4}\)

We rewrite the left sides as the square of a binomial and we add the numbers on the right side.

\(\left( x - \dfrac{3}{2} \right)^2 = \dfrac{16}{4} + \dfrac{9}{4} \)

\(\left( x - \dfrac{3}{2} \right)^2 = \dfrac{25}{4} \)

Now we apply the rule: If \(x^2 = k\), then \(x = \pm \sqrt k\) to get:

\( x - \dfrac{3}{2} = \pm \sqrt{\dfrac{25}{4}} \)

\( x - \dfrac{3}{2} = \pm \dfrac{5}{2} \)

We separate the equation into two parts:

\( x - \dfrac{3}{2} = \dfrac{5}{2} \) or \( x - \dfrac{3}{2} = -\dfrac{5}{2} \)

Add \(\dfrac{3}{2} \) to both sides in both equations:

\( x = \dfrac{8}{2} \) or \( x = -\dfrac{2}{2} \)

Reduce the fractions:

\( x = 4 \) or \( x = -1 \)

Answer 18

The step of completing the square by adding the square of half of the x-term coefficient to both sides only works if the coefficient of the x^2 term is 1. After we divided the equation by 3 on both sides, we did have a coefficient of 1 for the x^2 term, so we were able to complete the square that way.

Answer 19

i agree with you

Answer 20

i had previously solved the sum in that way but i deletedit