can you solve it explaining it for me? Especially the last factorised bit,,,It's :
3x(2x+3) -8 (2x+3)
= (2x-8)(2x+3)=0
I don't get the last line...if there are two (2x+3)...why isn't it written as (2x+3)^2?
Quadratics factorisation; Solve: 6x^2-7x-24=0?
2009-03-16 11:18 am
回答 (9)
2009-03-16 12:36 pm
✔ 最佳答案
(3x-----)(2x-----)(3x----8)(2x----3)
(3x - 8)(2x + 3)
(3x - 8) (2x + 3) = 0
x = 8/3 , x = - 3/2
2016-11-01 1:43 pm
hi Tracey, are you severe approximately this question? isn't it too trivial for dropping 5 useful factors on?? besides, right here we go: For the 1st term 6x^2, locate 2 factors such that their sum is under the 2nd term 7x. enable us to objective 3x and 2x and proceed as follows: (3x + .... )(2x + .... ) Now locate 2 factors such that their product is two and the manufactured from each with 3x and 2x respectively upload as much as 7x. (3x + 2)(2x + a million) word the positions of the two and a million. Reversing them does not yield the comparable undertaking.
2009-03-16 11:43 am
6x^2-7x-24=0
(2x+3)(3x-8) factor
2x+3=0 3x-8=0 zero-product property
2x=-3 3x=8 solve each equation
x=-3/2 x=8/3
x={-3/2,8/3} answer//
checking
6(-3/2)^2-7(-3/2)-24=0
6(9/4)+21/2-24=0
27/2+21/2-24=0
0=0 correct!
(2x+3)(3x-8) factor
2x+3=0 3x-8=0 zero-product property
2x=-3 3x=8 solve each equation
x=-3/2 x=8/3
x={-3/2,8/3} answer//
checking
6(-3/2)^2-7(-3/2)-24=0
6(9/4)+21/2-24=0
27/2+21/2-24=0
0=0 correct!
2009-03-16 11:35 am
reference:
3x (2x+3) - 8 (2x+3)
a (n) - b (n)
n (a - b)
we knew that n = 2x + 3
n (a - b)
(2x + 3) (a - b)
(2x + 3) (3x - 8)
That's it.
3x (2x+3) - 8 (2x+3)
a (n) - b (n)
n (a - b)
we knew that n = 2x + 3
n (a - b)
(2x + 3) (a - b)
(2x + 3) (3x - 8)
That's it.
2009-03-16 11:32 am
e.g.
8 + 16
= 1(8) + 2(8)
= 8(1 + 2)
6x^2 - 7x - 24
= 6x^2 + 9x - 16x - 24
= (6x^2 + 9x) - (16x + 24)
= 3x(2x + 3) - 8(2x + 3)
= (2x + 3)(3x - 8)
8 + 16
= 1(8) + 2(8)
= 8(1 + 2)
6x^2 - 7x - 24
= 6x^2 + 9x - 16x - 24
= (6x^2 + 9x) - (16x + 24)
= 3x(2x + 3) - 8(2x + 3)
= (2x + 3)(3x - 8)
2009-03-16 11:30 am
They aren't written as (2x+3)^2 because they are added (with coefficients), and not multiplied.
The same way 2x+2y isn't 2^2 (despite the fact there are two 2-s), but 2(x+y).
As for 6x^2-7x-24:
Quadratic formula:
x1 = (-b+sqrt(b^2-4ac))/(2a)
x2 = (-b-sqrt(b^2-4ac))/(2a)
a=6, b=-7, c=-24
sqrt(b^2-4ac) = sqrt(49+576) = 25
x1 = (7+25)/(2*6) = 32/12 = 8/3
x2 = (7-25)/12 = -18/12 = -3/2
So your factorization is: (3x-8)(2x+3)=0
The same way 2x+2y isn't 2^2 (despite the fact there are two 2-s), but 2(x+y).
As for 6x^2-7x-24:
Quadratic formula:
x1 = (-b+sqrt(b^2-4ac))/(2a)
x2 = (-b-sqrt(b^2-4ac))/(2a)
a=6, b=-7, c=-24
sqrt(b^2-4ac) = sqrt(49+576) = 25
x1 = (7+25)/(2*6) = 32/12 = 8/3
x2 = (7-25)/12 = -18/12 = -3/2
So your factorization is: (3x-8)(2x+3)=0
2009-03-16 11:29 am
It is the rule:
x(a+b)+ - y (a+b)
= (x- + y)(a+b)
(a+b) is a common factor.
x(a+b)+ - y (a+b)
= (x- + y)(a+b)
(a+b) is a common factor.
2009-03-16 11:25 am
No if there are two (2x+3) is not a square bec its not multiplied together , its subtraction
so treat it like another factorization eg xy - zy---> (x-z)y
so treat it like another factorization eg xy - zy---> (x-z)y
2009-03-16 11:24 am
because the is a minus between them, you cant only square if the two brackets are being multiplied.
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