Question 162857This question is from textbook
: x^2=2x-5
Should I layout this problem as:
x^2-2x-5=0
if so I followed through with the quadratic formula then from there I got
x = 2+24/2 = 26/2 = 13 and x = 2-24/2= -22/2= -11
so my solutions would be 13 and -11.
Please any help looking over would be appreciated.
This question is from textbook
Found 2 solutions by orca, joecbaseball:
Answer by orca(409)   (Show Source):
You can put this solution on YOUR website! The first step was wrong.
From x^2 = 2x - 5, we have
x^2 -2x + 2x - 5 -2x (subtract 2x from each side)
x^2 -2x = -5 (after combine like terms on the right side)
x^2 - 2x + 5 = -5 + 5 (add 5 to both sides)
x^2 - 2x + 5 = 0 (after combine like terms on the right side)
Answer by joecbaseball(37) (Show Source):
You can put this solution on YOUR website! Since I do not have any information on the grade level you are in, I will assume that you wrote the problem correctly, although the answer for the problem that you entered will have an imaginary answer:
You stated the problem as being x^2 = 2x – 5.
If that is correct, we set the equation to zero by adding –(2x-5) to both sides, or, maybe more simply, by adding (-2x) and adding 5 to both sides.
This gives you x^2 – 2x + 5 = 0.
We now must use the quadratic formula.
X=[ -b ± sqrt (b^2 – 4ac)] / 2a where a = 1, b = -2, and c = 5 from our equation.
This gives us:
X = [+2 ± sqrt (4 – 4(1)(5)] / 2
Which gives us….
X = [2 ± sqrt (-16)] / 2
Which gives us….
X = [2 ± 4i] / 2
Which gives us…
X = 1 ± 2i.
If you are not working with imaginary numbers yet, check the signs of the problem that you submitted. If the 5 is positive, then that would eliminate the imaginary answer.
Good luck.
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