solve 2x^2-5x=12 show work?
回答 (6)
2010-03-24 2:49 pm
✔ 最佳答案
2x^2 - 5x = 122x^2 - 5x - 12 = 0
(x - 4)(2x + 3) = 0
x = 4 or x = -3/2
參考: I have taught math for over 40 yr.
2010-03-24 10:18 pm
2x^2 - 5x = 12
2x^2 - 5x - 12 = 0
2x^2 + 3x - 8x - 12 = 0
(2x^2 + 3x) - (8x + 12) = 0
x(2x + 3) - 4(2x + 3) = 0
(2x + 3)(x - 4) = 0
2x + 3 = 0
2x = -3
x = -3/2 (-1.5)
x - 4 = 0
x = 4
â´ x = -3/2 (-1.5), 4
2x^2 - 5x - 12 = 0
2x^2 + 3x - 8x - 12 = 0
(2x^2 + 3x) - (8x + 12) = 0
x(2x + 3) - 4(2x + 3) = 0
(2x + 3)(x - 4) = 0
2x + 3 = 0
2x = -3
x = -3/2 (-1.5)
x - 4 = 0
x = 4
â´ x = -3/2 (-1.5), 4
2010-03-24 9:56 pm
2x^2 - 5x=12
2x^2 - 5x - 12=0
(2x +3) (x - 4)=0
so, x=-3/2 or x=4
hope this helps. =]
2x^2 - 5x - 12=0
(2x +3) (x - 4)=0
so, x=-3/2 or x=4
hope this helps. =]
2010-03-24 9:53 pm
2x^2-5x-12=0
(2x-3)(x-4)=0 x=4,3/2
(2x-3)(x-4)=0 x=4,3/2
2010-03-24 9:51 pm
2x^2 - 5x - 12 = 0
Check for solutions: b^2 - 4ac = 25 + 96 = 121 > 0, so there are two solutions.
Use the quadratic formula, [-b +- sqrt(b^2-4ac)]/2a to find the two solutions. They are -1.5 and 4.
Close. ;)
Check for solutions: b^2 - 4ac = 25 + 96 = 121 > 0, so there are two solutions.
Use the quadratic formula, [-b +- sqrt(b^2-4ac)]/2a to find the two solutions. They are -1.5 and 4.
Close. ;)
2010-03-24 9:49 pm
2x^2-5x=12
2x^2-5x-12 =0
5+-(sqrt)(25+96) / 4
5+- 11 / 4
16/4 = 4
-6/4 = -3/2
2x^2-5x-12 =0
5+-(sqrt)(25+96) / 4
5+- 11 / 4
16/4 = 4
-6/4 = -3/2
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