Help??!! ALGEBRA 1?

x^2 - 6x - 16 = 0

add 16 to both sides

x^2 - 6x = 16

now, take the -6 and divide it by 2 and then square it. add that numer to both sides of the equation

x^2 - 6x + 9 = 16 + 9

factor the left side (it is now a perfect square trinomial)

(x - 3)^2 = 25

take the square root of both sides

x - 3 = 5 and -5

add 3 to both answers

x = 8 and x = -2

There you go!

To solve this question by completing the square..

1. Look at the x^2. See if it has any numerals in front, e.g. 2x^2, 3x^2 etc. If there is, divide the whole equation by the numeral first.

2. If there isn't, look at the 6x. To complete a square, you wouldn't want to do anything to the x. In order to get the complete square term, divide the numeral in front of x by 2. In this case, you should get (x-3)^2. The negative or postive sign in the (x-3)^2 is due to the -6x. So if it was a 6x, it will be (x+3)^2 instead.

3. To make (x-3)^2 back to the equation in the question, you expand (x-3)^2 to get x^2-6x+9. For the equation in the question to still have -16, you have to minus the 9 in (x^2-6x+9). Therefore, you need to minus 25 from (x-3)^2.

4. Then you will get (x-3)^2 - 25 = 0. Then solve and get [(x-3) - 5][(x-3)+5] = 0. Therefore, x-8 = 0 or x+2 = 0. And, x=8 or x=-2.

I hope it helped. Good luck!

take out the x term (6) and divide by 2 which comes out to be 3 then square that which gets you 9 and then put 16 on the other side so it looks like this x^2-6x+9=16 then factor the left side (x-3)^2 = 16 but remember you added 9 to one side so you have to add 9 to the other so you have (x-3)^2 = 16 + 9 so you have (x-3)^2 = 25 then take the the square root of both sides but if you take the square root of a square the answer can be positive or negative so you get x - 3 = + or - 5 then add 3 to -5 and +5 to get your answers which are x = -2, 8

x^2-6x-16=0

(x-8)(x+2)=0 ; by factoring from above equation

So

x = 8 or x = -2

because any of these will cause the factors to be 0

x=8