I came up with:
+/- (sqrt14)/9
2007-12-14 21:16:38 · 9 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
x^2 = 49/9
x = +/- 7/3
7^2 = 49 and 3^2 = 9
2007-12-14 21:20:12 · answer #1 · answered by zi_xin 5 · 2⤊ 0⤋
Your pal is right 6x^2 - 2x = 0 x is a time-honored factor, so take an x out x(6x - 2) = 0 Now, set the two motives equivalent to 0 x = 0 6x - 2 = 0 treatment for x x = 0 6x - 2 = 0 6x = 2 x = 2/6 x = a million/3 The x-intercepts would be 0 and a million/3, like your chum pronounced. EDIT: specific, you're proper, a is 6 and b is -2, purely option x = -b/2a x = -(-2)/2(6) x = 4/12 x = a million/3, that's the x-coordinate of the vertex Now, option this into the equation to locate the y-coordinate. 6x^2 - 2x = y 6(a million/3)^2 - 2(a million/3) = y 6(a million/9) - (2/3) = y 2/3 - 2/3 = y 0 = y, the y-coordinate is 0 for this reason, the vertex is (a million/3 , 0)
2016-11-27 01:50:06 · answer #2 · answered by gravitt 4 · 0⤊ 0⤋
9x²=49
x²=49/9
x=√(49/9)
x=± 7/3
2007-12-14 21:22:41 · answer #3 · answered by Murtaza 6 · 1⤊ 0⤋
x=+-7/3
2007-12-14 21:29:50 · answer #4 · answered by rAvEn JoHn 2 · 0⤊ 0⤋
subtract 49 from both sides
9x^2 - 49 = 0
(3x+7)(3x-7) = 0
tada
2007-12-14 21:20:52 · answer #5 · answered by tom4bucs 7 · 2⤊ 0⤋
how could this NOT be 7/3..?
x^2 = 49/9..duh..!
2007-12-14 21:24:23 · answer #6 · answered by c0cky 5 · 0⤊ 0⤋
9x^2=49
x^2=49/9
x=(+/-)sqrt(49/9)
x=(+/-)7/3
2007-12-14 21:34:38 · answer #7 · answered by 315 1 · 0⤊ 0⤋
you have two answers:
x1= (sqrt(49 / 9))
x2= -(sqrt(49 / 9))
2007-12-14 21:57:30 · answer #8 · answered by Peyman 2 · 0⤊ 0⤋
the x's i got is 7/3 and -7/3...im not sure though.... i factored it out (3x+7)(3x-7)
2007-12-14 21:21:07 · answer #9 · answered by *just curious* 2 · 0⤊ 0⤋