quadratics: 3(x-2)^2 + 4 = 52, 1/4(x+1)^2 - 16 = 0

[belle<3]

ok so I am doing my algebra 2 hw yes. but how do I do these two? they are different.

#1. 3(x-2)^2 + 4 = 52
3(x-2)^2 + 4 -52 = 52 -52
3(x-2)^2 -48 = 0

#2. 1/4(x+1)^2 - 16 = 0
1/4(x^2+2x+1) - 16 = 0

I think I use the quadratic formula, which the other problems did not require. I know that the problem as to be equal to zero to work. This shows as far as I got on both. I know that x = -b + or - the sq. root of b^2 - 4ac all over 2a. But what do I do next on these.

 

[Mrspi]

Re: quadratics?

ok so I am doing my algebra 2 hw yes. but how do I do these two? they are different.

#1. 3(x-2)^2 + 4 = 52
3(x-2)^2 + 4 -52 = 52 -52
3(x-2)^2 -48 = 0


#2. 1/4(x+1)^2 - 16 = 0
1/4(x^2+2x+1) - 16 = 0


I think I use the quadratic formula, which the other problems did not require. I know that the problem as to be equal to zero to work. This shows as far as I got on both. I know that x = -b + or - the sq. root of b^2 - 4ac all over 2a. But what do I do next on these.

#1 contains a nice perfect square involving x. Let's see if we can get that on one side by itself, and then take the square root of both sides:

3(x - 2)[sup:1qclphwk]2[/sup:1qclphwk] + 4 = 52

Subtract 4 from both sides:

3(x - 2)[sup:1qclphwk]2[/sup:1qclphwk] = 48

Divide both sides by 3:

(x - 2)[sup:1qclphwk]2[/sup:1qclphwk] = 16

Take the square root of both sides:

x - 2 = + 4

Add 2 to both sides:

x = 2 + 4

x = 2 + 4, or x = 2 - 4
x = 6 or x = -2

See if you can apply the same approach to #2.