# Thread: [SOLVED] Completed Square Form

1. ## [SOLVED] Completed Square Form

I can't get the write answers for the following, can you please solve them step by step:

Factorize this by using the completed square method

$\displaystyle 12x^2+x-6$

Using the completed square form, find the least or greatest value of the following, and the value of x for which this occurs

$\displaystyle 4+6x=-x^2$

Thanks!

2. Originally Posted by unstopabl3
I can't get the [right] answers for the following....

Thank you!

3. a)
12x^2+x-6
12(x^2+x/12)-6
12{(x+1/24)^2+(1/24)(-1/24)}-6
12{(x+1/24)^2-1/576}-6
[tex]{12(x+1/24)^2+12(-1/576)}-6
{12(x+1/24)^2-1/48}-6
{12(x+1/24)^2-1/48-6(48/48)}
{12(x+1/24)^2-289/48}

From here onward I get stuck. Please tell me if I've made a mistake and solve from here onwards. Thanks.

b)

4+6x-x^2
-1x^2+6x+4
-{(x^2+6x)}+4
-{(x+6)^2}+4
-{(x+3)^2+3(-3)}+4
-{(x+3)^2-9}+4
-(x+3)^2+9+4
-(x+3)^2+13

when put x = 3
then f(x) = 13

The answer is correct but is this correct way to do it or is there another simpler method?

Thanks!

4. Originally Posted by unstopabl3

$\displaystyle 12x^2+x-6$

$\displaystyle 4+6x=-x^2$

where for the first equation: a=12 , b=1 , c=-6

for the second equation: a=1 b=6, c=4

5. Thanks but the question demands it that I solve it by using the completing square method

6. Originally Posted by unstopabl3
a)
12x^2+x-6
12(x^2+x/12)-6
12{(x+1/24)^2+(1/24)(-1/24)}-6
12{(x+1/24)^2-1/576}-6
[tex]{12(x+1/24)^2+12(-1/576)}-6
{12(x+1/24)^2-1/48}-6
{12(x+1/24)^2-1/48-6(48/48)}
{12(x+1/24)^2-289/48}
12((x+1/24)^2-289/576)
12((x+1/24)^2-(17/24)^2) <<<<<<<<< difference of squares!
12(x+1/24+17/24)(x+1/24-17/24))
12(x+3/4)(x-2/3) <<<<<<<<< now split the factor 12 into 3 * 4 and put them into the appropriate bracket
(4x+3)(3x-2)

b)

4+6x=-x^2 <<<<<<<<< according to the wording of the problem
x^2+6x+4 = 0
(x^2+6x+9-9)+4=0
(x+3)^2-5=0
(x+3)^2-(sqrt(5))^2=0 <<<<<<<<<<< difference of squares

Factor the difference.
Aproduct equals zero if one factor equals zero.
Solve each equation for x.

Thanks!
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