How do you work out this Quadratic:: (x+3)(x-4)=18?
回答 (6)
2008-03-15 6:22 pm
✔ 最佳答案
x² - x - 12 = 18x² - x - 30 = 0
( x - 6 ) ( x + 5 ) = 0
x = 6 , x = - 5
2008-03-16 1:02 am
(x + 3)(x - 4) = 18
x^2 + 3x - 4x - 12 = 18
x^2 - x - 12 - 18 = 0
x^2 - x - 30 = 0
(x - 6)(x + 5) = 0
x - 6 = 0
x = 6
x + 5 = 0
x = -5
â´ x = 6 , -5
x^2 + 3x - 4x - 12 = 18
x^2 - x - 12 - 18 = 0
x^2 - x - 30 = 0
(x - 6)(x + 5) = 0
x - 6 = 0
x = 6
x + 5 = 0
x = -5
â´ x = 6 , -5
2008-03-16 1:02 am
(x+3)(x-4)=18
x^2+3x-4x-12=18
x^2-x-30=0
(x-6)(x+5)=0
x=6 or x=-5
hope that helped
x^2+3x-4x-12=18
x^2-x-30=0
(x-6)(x+5)=0
x=6 or x=-5
hope that helped
2008-03-16 12:59 am
you did ask how.
multiply out the left side. Move the 18 to the left and set equation =0. Combine terms. Use quadratic equation to solve.
multiply out the left side. Move the 18 to the left and set equation =0. Combine terms. Use quadratic equation to solve.
2008-03-16 1:06 am
First, multiply back to polynomial form:
x^2 -x -12 = 18
Then, subtract both sides by 18, so that you leave zero on the right side of the equal sign:
x^2 - x -30 = 0
Now, factor once again (reverse foil):
(x + 5)(x - 6) = 0
To satisfy :
x = -5 or x = 6
Plug back in to original equation to see if this solution set works!
x^2 -x -12 = 18
Then, subtract both sides by 18, so that you leave zero on the right side of the equal sign:
x^2 - x -30 = 0
Now, factor once again (reverse foil):
(x + 5)(x - 6) = 0
To satisfy :
x = -5 or x = 6
Plug back in to original equation to see if this solution set works!
2008-03-16 1:02 am
You first multiply 4 and 3 which equals 12. Then you subtract it from 18 which makes 6. Then you so you are six short. Then you put these numbers in:
(3+3)(7-4)=18 and that is the answer.
(3+3)(7-4)=18 and that is the answer.
參考: My Brain
收錄日期: 2021-05-01 10:14:46
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