I need to use the square root property to solve this problem?

not correct

(c-2)^2 =c^2 -4c +4 =12

c^2 -4c -8 =0

solution

(c - 2)^2 =12

c-2 =sqrt(12) = 3.464

c =3.464 +2 =5.464

Doesn't seem quite right - as you've seen, you'd need c to be the square root of -4, which can be done, but then wouldn't solve the original equation.

I would take the square root first:

(c-2) = SqRoot(12)

c = SqRoot(12)+2

I think in your attempt you didn't square the bracket quite right:

(c-2)^2 = (c-2)(c-2) = c^2 - 2c - 2c + 4 = c^2 - 4c + 4

Set that equal to 12:

c^2 - 4c + 4 = 12

c^2 - 4c - 8 = 0

Then use the quadratic formula to solve for c:

c = [-(-4) +/- SqRoot ((-4)^2)-4*1*(-8)) ] / 2(1)

c = [4 +/- SqRoot (16+32)] / 2 = [4 +/- SqRoot (48)] / 2 = 2 +/- SqRoot (12) as above.

(Note in the first answer above I hadn't considered the negative root. Whether or not you need to depends on the level of maths you're at.

Hope that helps.

Your expansion of (c-2)^2 is wrong . It should be c^2-4c+4.

There is another way . Take the sqroot

c-2=plus minus root of 12 .

Root of 12 is 2 x root of 3..

c= 2 plus/minus (2 multiplied by root of 3).

You should get the same answer by solving c^2-4c-8=0.

This is what you need to do:

(1) (c - 2)^2 = 12

Take the sqrt of both sides:

=> (c - 2) = +/- sqrt(12) = +/-sqrt(2*2*3) = +/- 2sqrt(3)

Add 2 to both sides:

=> c = +/- 2 sqrt(3) + 2 or

=> c = [2 sqrt(3) - 2] or c = [2 + 2 sqrt(3)]

(2) c^2 + 8 = 12

=> c^2 = 4

=> c = +/- sqrt(4)

=> c = 2 or - 2

not quite, (c-2)^2 means (c-2)*(c-2) then to simplify you'd use foil method to end up with c^2-4c+4 but in this question I think you would take square root of both sides to give you c-2=sqrt(12) then add 2 to both sides to end up with c=sqrt(12)+2

It is not correct.

c - 2 = ± 2√3

c = 2 ± 2√3

(5k-a million)^2 = 25k^2 -10k + a million 25*ok^2 - 10k + a million - 23 = 25k^2 - 10*ok - 22 = 0 a=25 , b= -10 , c = -22 quadratic formula: (-b +/- sqrt(b^2-4ac)/2a the +/- is plus or minus! once you upload it provides you you with the first root and once you subtract it provides you you with the 2d root. wish this facilitates