How do you factor 2x^2-3x-18?
回答 (5)
2010-07-17 4:27 am
✔ 最佳答案
Use the quadratic formula and sub in all of your values. Your answers should be 3.84 and -2.34
2010-07-17 11:37 am
2x^2-3x-18 = 2{ x^2 - 3/2 x -9} =
=2{ (x^2 - 3/2 x) -9}
= 2{ (x^2 -3/ 2 x +9/16) - 9 -9/16}
= 2{ (x - 3/ 4)^2 -(sqrt 151/16)^2}
= 2 { x - 3/4 +sqrt(151)/4} { x - 3/4 - sqrt(151)/4} ....................Ans
=2{ (x^2 - 3/2 x) -9}
= 2{ (x^2 -3/ 2 x +9/16) - 9 -9/16}
= 2{ (x - 3/ 4)^2 -(sqrt 151/16)^2}
= 2 { x - 3/4 +sqrt(151)/4} { x - 3/4 - sqrt(151)/4} ....................Ans
2010-07-17 11:33 am
You can always use the quadratic formula, with a=2; b=-3; and c=-18
= {3 +/- sqrt(153)} / 4
= (1/4)(3 +/- 3sqrt(17))
The two roots are:
= (3/4){1 + sqrt(17)}
= (3/4){1 - sqrt(17)}
= {3 +/- sqrt(153)} / 4
= (1/4)(3 +/- 3sqrt(17))
The two roots are:
= (3/4){1 + sqrt(17)}
= (3/4){1 - sqrt(17)}
2010-07-17 11:26 am
2010-07-17 11:25 am
It doesn't factor using integers.
You can tell because b^2 - 4ac = (-3)^2 - 4(2)(-18) = 9 - (-144) = 153, which is not a perfect square.
You can tell because b^2 - 4ac = (-3)^2 - 4(2)(-18) = 9 - (-144) = 153, which is not a perfect square.
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