Add the -25 to the other side...
2x^2-8x+25=0,
Now use the quadratic formula to solve for x...
Hope this helps
This is my question 2x^2-8x=-25
I cant seem to figure out how to solve for x?
4 answers
I was told to start with
x^2-8x+16=7/2 so where do i go from there?
x^2-8x+16=7/2 so where do i go from there?
2x^2 - 8x =-25
divide each term by 2
using completing the square,
x^2 - 4x = -25/2
take 1/2 the coefficient of the x term, square it, then add it to each side
x^2 - 4x +4 = -25/2 +4
(x-2)^2 = -17/2 or -34/4 , (I anticipated to take √ of both sides)
x -2 = ± √-34/2 = ± i√34/2
x = 4/2 ± i√34/2 = (4 ± i√34)/2
or ... by the formula
2x^2 - 8x + 25 = 0
x = (8 ± √-136)/4
= (8 ± 2√-34)/2
= (4 ± i√34)/2
divide each term by 2
using completing the square,
x^2 - 4x = -25/2
take 1/2 the coefficient of the x term, square it, then add it to each side
x^2 - 4x +4 = -25/2 +4
(x-2)^2 = -17/2 or -34/4 , (I anticipated to take √ of both sides)
x -2 = ± √-34/2 = ± i√34/2
x = 4/2 ± i√34/2 = (4 ± i√34)/2
or ... by the formula
2x^2 - 8x + 25 = 0
x = (8 ± √-136)/4
= (8 ± 2√-34)/2
= (4 ± i√34)/2
I was told to start with
x^2-8x+16=7/2 so where do i go from there? That's after / by 2.
x^2-8x+16=7/2 so where do i go from there? That's after / by 2.