use factors to solve 2n squared + 8n -24 =0?
2008-03-24 2:58 pm
I know the answer is -6,2 but can someone show me step by step how to solve this simply please!
回答 (11)
2008-03-24 3:02 pm
✔ 最佳答案
n^2 + 4n - 12 = 0Now you're looking for two numbers whose product is -12, and whose sum is +4.
(n+6) (n-2) = 0
2008-03-24 3:10 pm
= 2n² + 8n - 24
= 2(n² + 4n - 12)
= 2(n + 6)(n - 2)
n + 6 = 0, n = - 6
n - 2 = 0, n = 2
Answer: n = - 6, 2
Proof (where n is - 6):
2n² + 8n = 24
2(- 6²) + 8(- 6) = 24
2(36) - 48 = 24
72 - 48 = 24
24 = 24
Proof (where n is 2):
2(2²) + 8(2) = 24
2(4) + 16 = 24
8 + 16 = 24
24 = 24
= 2(n² + 4n - 12)
= 2(n + 6)(n - 2)
n + 6 = 0, n = - 6
n - 2 = 0, n = 2
Answer: n = - 6, 2
Proof (where n is - 6):
2n² + 8n = 24
2(- 6²) + 8(- 6) = 24
2(36) - 48 = 24
72 - 48 = 24
24 = 24
Proof (where n is 2):
2(2²) + 8(2) = 24
2(4) + 16 = 24
8 + 16 = 24
24 = 24
2008-03-25 5:16 pm
n² + 4n - 12 = 0
(n + 6)(n - 2) = 0
n = - 6 , n = 2
(n + 6)(n - 2) = 0
n = - 6 , n = 2
2008-03-24 8:43 pm
2n^2 + 8n - 24 = 0
(2n + 12)(n - 2) = 0
This means:
(2n + 12) = 0
and
(n - 2) = 0
2n + 12 = 0
2n = -12
n = -6
OR
n -2 = 0
n = 2
(2n + 12)(n - 2) = 0
This means:
(2n + 12) = 0
and
(n - 2) = 0
2n + 12 = 0
2n = -12
n = -6
OR
n -2 = 0
n = 2
2008-03-24 4:46 pm
2n^2 + 8n - 24 = 0
(2n - 4)(n + 6) = 0
2n - 4 = 0
2n = 4
n = 4/2
n = 2
n + 6 = 0
n = -6
∴ n = 2 , -6
(2n - 4)(n + 6) = 0
2n - 4 = 0
2n = 4
n = 4/2
n = 2
n + 6 = 0
n = -6
∴ n = 2 , -6
2008-03-24 3:10 pm
2n*n+8n-24=0=
2*(n*n+4n-12)=
n*n+4n-12
factors of 12: [(12,1),(6,2),(3,4)]
since the 0th power term is neg (-12), this implies we factor into a positive and a negative.
So out of our 3 factors of 12, figure out which one has a difference of 4.
1-12=-11
2-6=-4
3-4= -1
12-1=11
6-2=4
4-3=1
Ok, so (6) and (-2).
that leaves us with
(n+6)*(n-2)=2
setting each part to 0 and solving
n+6=0
n-2=0
thus m={-6,2}
2*(n*n+4n-12)=
n*n+4n-12
factors of 12: [(12,1),(6,2),(3,4)]
since the 0th power term is neg (-12), this implies we factor into a positive and a negative.
So out of our 3 factors of 12, figure out which one has a difference of 4.
1-12=-11
2-6=-4
3-4= -1
12-1=11
6-2=4
4-3=1
Ok, so (6) and (-2).
that leaves us with
(n+6)*(n-2)=2
setting each part to 0 and solving
n+6=0
n-2=0
thus m={-6,2}
2008-03-24 3:07 pm
This is how you do it.
2n^2+8n-24=0
2n^2+12n-4n+24=0 (you split the 8 as 12 and -4)
2n(n+6)-4(n+6)=0
(n+6)(2n-4)=0
(n+6)2(n-2)=0
2(n+6)(n-2)=0
n+6=0
n=-6
n-2=0
n=2
2n^2+8n-24=0
2n^2+12n-4n+24=0 (you split the 8 as 12 and -4)
2n(n+6)-4(n+6)=0
(n+6)(2n-4)=0
(n+6)2(n-2)=0
2(n+6)(n-2)=0
n+6=0
n=-6
n-2=0
n=2
2008-03-24 3:07 pm
Use the form;
AX2 + BX + C = 0
then;
x = (-b +/- sqrt(b^2 - 4ac))/2a
so plug and chug
x = (-8 +/- sqrt(8^2 - 4(2)(-24)) / 2(2)
x = (-8 +/- sqrt(64 + 192) / 4
x = (-8 +/- sqrt(256)) / 4
x = (-8 +/- 16) / 4
x = -24/4 & 8/4
x = -6 & 2
AX2 + BX + C = 0
then;
x = (-b +/- sqrt(b^2 - 4ac))/2a
so plug and chug
x = (-8 +/- sqrt(8^2 - 4(2)(-24)) / 2(2)
x = (-8 +/- sqrt(64 + 192) / 4
x = (-8 +/- sqrt(256)) / 4
x = (-8 +/- 16) / 4
x = -24/4 & 8/4
x = -6 & 2
2008-03-24 3:04 pm
2n^2 +8n -24 =0
n^2 +4n -12 =0
(n-2)(n+6) =0
n-2=0, n+6 =0
n=2,-6
n^2 +4n -12 =0
(n-2)(n+6) =0
n-2=0, n+6 =0
n=2,-6
2008-03-24 3:03 pm
2x^2 + 8n - 24 = 0 factors to
(2n - 4)(n + 6) = 0
Set both factors equal to 0
2n - 4 = 0
n + 6 = 0
Solve for n
2n = 4
n = 2
n + 6 = 0
n = -6
(2n - 4)(n + 6) = 0
Set both factors equal to 0
2n - 4 = 0
n + 6 = 0
Solve for n
2n = 4
n = 2
n + 6 = 0
n = -6
收錄日期: 2021-05-01 10:15:12
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