I'm having a problem completing a quadratic equation by completing the square?

Question

Could someone please explain step by step how I could solve the following equation by completing the square: k^2-8k+12=0 . I'm getting stuck when it comes to factoring it out. So far I have k^2-8k+12=0
k^2-8k=-12
k^2-8k+16=-12+16
(k+4)^2=4
This is where I get stuck at do I need to further square k+4. I have the answer to the question but the answer I figures out doesn't match what the correct answer is supposed to be.

Answer 1

k^2-8k+12=0
k^2-8k=-12 ok
k^2-8k+16=-12+16 ok
(k+4)^2=4 wrong
(k - 4)^2 = 4

Now take the square root:
k - 4 = +/- 2
k = 4 + 2 or 4 -2
= 6 or 2.

The factors (k - 6)(k - 2) check this out.

Answer 2

You are nearly correct so far, except that in the last step you wrote k^2 - 8k + 16 = (k + 4)^2, when in fact it is equal to (k - 4)^2. Now you just take (k - 4)^2 = 4 and say that k - 4 = +/- sqrt(4) ==> k = 4 +/- 2 = 2 or 6.

Answer 3

U can do it this way:
k^2-8k+12=0
or,k^2-6k-2k+12=0( u have to replace -8k by staffs that multiply to form 12K^2 n obviously add up to -8k that is -6k and -2k)
or,k(k-6)-2(k-6)=0
or,(k-6)(k-2)=0
Either k-6=0 or k-2=0
So, k=6 or k=2

Answer 4

The Problem is in the last step.
"k^2-8k+16=-12+16
(k+4)^2=4"
As it is -8k,shouldn't in the next step it should be [(k-4)^2=4] rather than [(k-4)^2=4].
Thereby, the next steps should be,
k-4=+-2 [after square rooting]
k=6 or k=2. [as square rooting left us with +2 & -2]

Bye Pal

Answer 5

k^2-8k+12=0
k(k-2)-6(k-2)+12 = 0
(k-6)(k-2) = 0
therefore k = 6 or 2

Answer 6

It looks like it should be (k-4)^2 on the left, not (k+4)^2.

Then you can take square root and get

k-4 = 2
k-4 = -2

or k = 2, 6.

Answer 7

Step (a million) ax^2 + bx + c = 0 If a isn't 0, we are able to divide by using a, to get (2) x^2 + (b/a)x + (c/a) = 0 /a = 0 Subtract (c/a) from the two facets (3) x^2 + (b/a)x = -(c/a) upload to the two facets: (4) x^2 + (b/a)x + (b^2/(4a^2))= -(c/a) + (b^2/(4a^2)) The left area is a suited sq. (5) x^2 + (b/a)x + (b^2/(4a^2))= (x + b/(2a))^2 So we've (6) (x + b/(2a))^2 = -(c/a) + (b^2/(4a^2)) = (b^2 - 4ac ) / (4a^2) Take the sq. root of the two facets (7) (x + b/(2a)) = +/-sqrt[ (b^2 - 4ac ) ] / (2a) Subtract b/(2a) from the two facets (8) x = +/-sqrt[ (b^2 - 4ac ) ] / (2a) - b/(2a) x = [-b +/-sqrt[ (b^2 - 4ac ) ] ] / (2a) This completes the answer for X.

Answer 8

(k)^2-2(k)(4)+12=0
by adding and subtracting 4^2,we get
k^2-2[k][4]+4^2-4^2+12=0
(k-4)^2=4^2-12
[k-4]^2=4
k-4= +-2
k= +-2+4
k=6 or k=2

Answer 9

k^2-8k+12=0
k^2-2k-6k+12=0.......................middle term break
k(k-2)-6(k-2)=0
(k-2)(k-6)=0.......................by taking (k-2) common

Therefore either k = 2 or k = 6
as simple as that