solve for x: 6x^2+7x-3?
回答 (10)
2009-11-06 1:49 pm
✔ 最佳答案
6x^2+7x-3(2 x+3) (3 x-1)=0
2x+3=0 3x-1=0
2x=-3 3x=1
x=-3/2 x=1/3
x=-3/2,1/3 answer//
2009-11-06 1:44 pm
Factorise: (3x-1)(2x+3)=0
x=1/3 or -3/2
x=1/3 or -3/2
2016-12-15 5:58 pm
The discriminant is the fee of the expression below the unconventional sign interior the quadratic formula it particularly is b^2-(4ac) on the grounds which you're coping with a quadratic equation interior the variety ax^2+bx+c=0 you will locate the discriminant. subsequently a=5/6, b=-7 and c=-6/5. Plug the numbers into the discriminant, it particularly is b^2-(4ac) as above, to locate the fee of the discriminant. (i will go away that as much as you on the grounds which you're in a variety larger that Algebra a million.......) If the discriminant you detect is: detrimental - the concepts are complicated (there are no longer any actual concepts) corresponding to 0 - there is one actual answer powerful - there are 2 actual concepts.
2015-05-28 8:01 am
1/3
2009-11-07 6:11 am
6x^2 + 7x - 3
= 6x^2 + 9x - 2x - 3
= (6x^2 + 9x) - (2x + 3)
= 3x(2x + 3) - 1(2x + 3)
= (2x + 3)(3x - 1)
= 6x^2 + 9x - 2x - 3
= (6x^2 + 9x) - (2x + 3)
= 3x(2x + 3) - 1(2x + 3)
= (2x + 3)(3x - 1)
2009-11-06 3:39 pm
It is a hard question for me, sorry.
2009-11-06 1:53 pm
I am putting the expression equal to 0. Hope this is correct.
Using ac method
a = 6
c = -3
a times c = -18
Factors of -18 to give +7 (the middle number) when added = +9 - 2
Rearranging gives
6x*2 +9x - 2x -3 = 0
Bracketing
3x[2x + 3] - 1[2x + 3] = 0
Take out the common bracket gives
[2x + 3][3x - 1] = 0
Either 2x + 3 = 0 and x = -3/2
Or 3x -1 = 0 and x = 1/3
[If an answer is zero when 2 numbers are multiplied together one of the numbers must be equal to zero]
Using ac method
a = 6
c = -3
a times c = -18
Factors of -18 to give +7 (the middle number) when added = +9 - 2
Rearranging gives
6x*2 +9x - 2x -3 = 0
Bracketing
3x[2x + 3] - 1[2x + 3] = 0
Take out the common bracket gives
[2x + 3][3x - 1] = 0
Either 2x + 3 = 0 and x = -3/2
Or 3x -1 = 0 and x = 1/3
[If an answer is zero when 2 numbers are multiplied together one of the numbers must be equal to zero]
2009-11-06 1:52 pm
to find the roots plug the following coefficients into the quadratic formula :
a = 6
b = 7
c = -3
x = {-7 +/- [7^2 -4(6)(-3)]^1/2 } / 2(6)
x = {-7 +/- [49 + 72]^1/2 } / 12
x = {-7 +/- 11} / 12
x = 0.33333 , -1.5
a = 6
b = 7
c = -3
x = {-7 +/- [7^2 -4(6)(-3)]^1/2 } / 2(6)
x = {-7 +/- [49 + 72]^1/2 } / 12
x = {-7 +/- 11} / 12
x = 0.33333 , -1.5
2009-11-06 1:50 pm
6x^2+7x-3=0 - equate to zero
(3x - 1)(2x + 3)=0 - factor out
(3x - 1)=0 - equate each to zero
(2x + 3)=0
x = 1/3 , x = -3/2 - solve for x in each factor
(3x - 1)(2x + 3)=0 - factor out
(3x - 1)=0 - equate each to zero
(2x + 3)=0
x = 1/3 , x = -3/2 - solve for x in each factor
2009-11-06 1:45 pm
x= -6.39
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