Help with solving for X?

Question

I have to solve to following problem for X , but I am lost...

5(x-2)^2 = 3

Also this problem has me scratching my head ....

I have to solve this problem by either "completing the square" or using the "quadratic formula" Help !!!

x^2 = 5x +2

Do I set it up as x^2 -5x -2 = 0 and where do I go from there???

Answer 1

5(x-2)^2 = 3 divide both sides by 5
(x-2)^2=0.6 take square root of both sides
x-2=+/-√0.6
x=2+√0.6, 2-√0.6

x^2 = 5x +2 subtract 5x+2 from each side
x^2-5x-2=0
x=(5+/-√(25+8))/2
x=(5+/-√33)/2
x=2.5+√33, 2.5-√33

Answer 2

I'm no math genius (in 8th grade), but i think quadratic formula works. 5(x-2)^2=3. Expand to 5(x^2-4x+4)=3. Simplify. 5x^2-20x+17=0. Quadratic Formula. (-b+/- Squareroot of b^2-4ac)/2a. so... (20+/- Squareroot of 400-340)/10 so it becomes (20+/- Squareroot of 60)/10. Factor out squareroot of 60 into 2squareroot of 15. (20+/- 2squareroot of 15)/10. divide by 2 to get final answer of (10+/-squareroot of 15)/5. Hope that helps. Don't kill me if I'm wrong :D

Answer 3

I'm getting x=4.2/x, although I don't know if I did it right.
I did this:
5(x-2)^2=3, distributive property, (5x-10)^2=3, square,
25x^2-100=3, move 100, 25x^2=103, move 25, x^2=4.2, devide both sides by x to get x alone instead of x^2, x=4.2/x. If you need an answer, test x^2=4.2. I'm not sure what you'd do for the (5x-10)^2.

Answer 4

You foiled wrong.

(x-2)^2 = X^2 -4x +4

then.. you get

5X^2 - 20x + 20

subtract 3 you get

5X^2 - 20x+17

Then you would need to complete the square or do the quadratic formula.

Answer 5

if u use the quad formula,, you just pug in a b and c using your numbers and then solve. in this case a would be 1, b -5 and c -2. if u need the quad form look in th eback of ur textbook...