ok here are the 2 equations y=x^2-x and y=4-x. Ive gotton it downe to this:
4-x=x^2-x
+x +x
4=x^2-1
-4 -4
0=x^2-5
or
x^2-5=0
and now im stuck...any ideas?
2006-11-14 11:11:05 · 8 answers · asked by xombiecats 2 in Science & Mathematics ➔ Mathematics
When you added x to the right side, you should have gotten:
4 = x^2
Then take the square root of both sides to get
x = 2, -2
2006-11-14 11:13:49 · answer #1 · answered by Kilowatt 2 · 0⤊ 0⤋
ok. x^2-5 = 0.
to find x you would add 5 to the other side which would give you x^2=5 then take the square root of both sides and x = the square root of 5
2006-11-14 11:14:07 · answer #2 · answered by Anonymous · 0⤊ 0⤋
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2016-10-03 23:25:19 · answer #3 · answered by ? 4 · 0⤊ 0⤋
4-x = x^2 -x
+x +x
4= x^2
x= +2, -2
2006-11-14 11:15:43 · answer #4 · answered by maegical 4 · 0⤊ 0⤋
y = x^2 - x and y = 4 - x ... so...
x^2 - x = 4 - x ...
You are looking too far into this... just add 'x' to both sides...
x^2 = 4
x = 2 or x = -2
Hope this helps you!
2006-11-14 11:17:49 · answer #5 · answered by Anonymous · 0⤊ 0⤋
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2006-11-14 11:14:34 · answer #6 · answered by Brian F 1 · 0⤊ 1⤋
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2006-11-14 11:18:48 · answer #7 · answered by Anonymous · 0⤊ 1⤋