x+ √34-x^2 = 2
i'm really confused about this so if anyone can help then that would be really appreciated =)
so far i could only get it up to standard form (if thats what i'm suppose to do) and from there i'm clueless. help?
2007-05-19 07:13:15 · 5 answers · asked by over the air 3 in Science & Mathematics ➔ Mathematics
move x to the other side and then square both sides of the equation
√34-x^2 = 2 - x
(√34-x^2)^2 = (2 - x)^2
34 - x^2 = 4 - 4x + x^2
2x^2 - 4x -30 = 0
2(x^2 - 2x - 15) = 0
x^2 - 2x - 15 = 0
(x - 5)(x + 3) = 0
x = 5, x = -3
2007-05-19 07:18:02 · answer #1 · answered by Ana 4 · 0⤊ 1⤋
sqrt(34-x^2)=2-x
Before squaring you must be sure that both sides hve the same sign
so 2-x>=0 and x<=2
now
34-x^2=x^2-4x+4
2x^2-4x -30=0 and x= ((4+-sqrt(16+240))/4
x=5(not a solution)and x=-3 (only solution)
2007-05-19 15:51:49 · answer #2 · answered by santmann2002 7 · 0⤊ 0⤋
x=-3
2007-05-19 14:27:27 · answer #3 · answered by Anonymous · 0⤊ 0⤋
x+sqr(34) -x^2=2
Rewrite the equation:
-x^2 +x+ [sqr(34)-2]=0
Then apply the formula for solving the second digrre equation.
2007-05-19 14:21:34 · answer #4 · answered by ali j 2 · 0⤊ 0⤋
????
2007-05-19 14:19:58 · answer #5 · answered by Anonymous · 0⤊ 1⤋