solve the equation 2x^2-35x+98=0
(without a calc please)
2007-04-16 02:20:06 · 8 answers · asked by jenny 1 in Science & Mathematics ➔ Mathematics
2x² - 35c + 98 = 0
2x² - 28x - 7x + 98 = 0
2x(x - 14 - 7(x - 14) 0
)2x - 7)(x - 14)
- - - - - - -s-
2007-04-16 03:06:54 · answer #1 · answered by SAMUEL D 7 · 0⤊ 0⤋
Solve,
2x^2 - 35x + 98 = 0
(2x - 7)(x - 14) = 0
x = 7/2 or x = 14
Working...
98 = 2 x 49
98 = 2 x 7 x 7
98 = 14 x 7
2007-04-16 02:48:35 · answer #2 · answered by ideaquest 7 · 0⤊ 0⤋
2x² - 35x + 98 = 0
One way is to use formula:
[-b ± √(b² - 4ac)]/ 2a
[-(-35) ± √(35² - 4(2)(98)]/ 2(2)
[-(-35) ± √(1225 - 784)]/ 4
[-(-35) ± √(441]/ 4
[-(-35) ± 21]/ 4
(+35 ± 21)/ 4
+56/ 4 or 14/ 4
x = 14 or x = 3∙5
2007-04-16 03:15:46 · answer #3 · answered by Brenmore 5 · 0⤊ 0⤋
You don't want to expand it. One of the properties of 0 is that if a product of terms =0, at least one of the terms must =0. Therefore, the solutions of 3x(1 - x)(4x + 5) = 0 are x=0, (1-x)=0 which means x=1, or (4x+5)=0 which means that x= -5/4
2016-04-01 04:06:12 · answer #4 · answered by Anonymous · 0⤊ 0⤋
No probs.....
(2x - 7) (x - 14) = 0
x = 3.5 or x = 14
2007-04-16 02:25:13 · answer #5 · answered by Doctor Q 6 · 2⤊ 1⤋
2x^2-28x-7x+98
2x(x-14)-7(x-14)
(2x-7) (x-14)
2007-04-16 02:35:23 · answer #6 · answered by Anonymous · 0⤊ 0⤋
(2x-7) (x-14)
2007-04-16 02:34:00 · answer #7 · answered by Pvt. Andrew Malone 5 · 0⤊ 0⤋
2[x^2-17.5x+49]
2[x^2-14x-3.5x+49]
2[x{x-14}-3.5{x-14}]
2[{x-14}{x-3.5}]
=2(x-14)(x-3.5)
2007-04-16 02:23:24 · answer #8 · answered by nishit 2 · 0⤊ 1⤋