SOLUTION: My homework is :1) X2-2x-13=0 and 2) X2+12x-64=0. I am do use six steps of 1) move constant term to right side of equation
2) multiply each term in the equation by 4 times th
Question 864126: My homework is :1) X2-2x-13=0 and 2) X2+12x-64=0. I am do use six steps of 1) move constant term to right side of equation
2) multiply each term in the equation by 4 times the coefficient of the original x2 term.
3) Square the coefficient of the original x term and add it to both sides of the equation.
4) Take the square root of both sides
5) Set the left side of the equation equal to the positive square root of the number on the right side and solve for x.
6) Set the left side of the equation equal to the negative square root of the number on the right side of the equation and solve for x.
Please show me how these two problems will look worked out and then I should be able to look at these to see how to do them. Thank you. Answer by richwmiller(17219) (Show Source):
You can put this solution on YOUR website! Your instructions are wrong for completing the square.
You or someone else have presented these instructions before. Get rid of them.
Plus be careful : x is no the same as X
You or someone else have presented this error before.
x^2-2x-13=0
x^2-2x=13
(1/2*2)^2=1^2=1
x^2-2x+1=14
(x-1)^2=14
x-1=+-sqrt(14)
x=1+sqrt(14)
x=1-sqrt(14)
x^2+12x-64=0
This could also be factored
(x-4) (x+16) = 0
x=4 x=-16
complete the square
x^2+12x-64=0
x^2+12x=64
12/2=6
6^2=36
x^2+12x+36=100
(x+6)^2=100
x=-6-10=-16
x=-6+10=4
