quadratic formula n² +3n+2=0?
回答 (9)
2009-04-21 3:55 pm
✔ 最佳答案
n^2 + 3n + 2 = 0n^2 + 2n + n + 2 = 0
n(n + 2) + 1(n + 2) = 0
(n + 1)(n + 2) = 0
n = -1 ...OR... n = -2
2009-04-21 10:50 pm
(n+1)(n+2)=0
n=-1 & n=-2
n=-1 & n=-2
2009-04-22 3:35 am
This is easier done with factoring, but if you have to use the quadratic formula:
Refer to my Math Variety blog post on how to solve quadratic equations by using the quadratic formula and feel free to click on any ads/links of interest
http://mathvariety.blogspot.com/2009/04/solving-quadratic-equations-by-using.html
Check out my eHow article on how to solve quadratic equations by using the quadratic formula:
http://www.ehow.com/how_4816180_quadratic-equation-using-quadratic-formula.html
Also check out this link. It’s a calculator that will give you the roots (solutions) to any quadratic equation. Just enter values for a, b, and c.
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The quadratic formula is:
n = [-b +/- sqrt(b^2-4ac)]/(2a)
Here, a=1, b=3, c=2. We have:
n=[-3 +/- sqrt(3^2-4*1*2)]/(2*1)
n=[-3 +/- sqrt(9-8)]/2
n=[-3 +/- sqrt(1)]/2
n= -1 or n= -2
Refer to my Math Variety blog post on how to solve quadratic equations by using the quadratic formula and feel free to click on any ads/links of interest
http://mathvariety.blogspot.com/2009/04/solving-quadratic-equations-by-using.html
Check out my eHow article on how to solve quadratic equations by using the quadratic formula:
http://www.ehow.com/how_4816180_quadratic-equation-using-quadratic-formula.html
Also check out this link. It’s a calculator that will give you the roots (solutions) to any quadratic equation. Just enter values for a, b, and c.
http://www.1728.com/quadratc.htm
The quadratic formula is:
n = [-b +/- sqrt(b^2-4ac)]/(2a)
Here, a=1, b=3, c=2. We have:
n=[-3 +/- sqrt(3^2-4*1*2)]/(2*1)
n=[-3 +/- sqrt(9-8)]/2
n=[-3 +/- sqrt(1)]/2
n= -1 or n= -2
2009-04-21 11:14 pm
n= -2 and -1
2009-04-21 10:58 pm
first factorize it (hope you know)
3n can be split into 2 + 1
so, 0 = (n+2)(n+1)
n + 2 =0, n +1=0
n = -2 and n= -1
3n can be split into 2 + 1
so, 0 = (n+2)(n+1)
n + 2 =0, n +1=0
n = -2 and n= -1
2009-04-21 10:57 pm
n^2 + 3n + 2 = 0
n^2 + 2n + n + 2 = 0
(n^2 + 2n) + (n + 2) = 0
n(n + 2) + 1(n + 2) = 0
(n + 2)(n + 1) = 0
n + 2 = 0
n = -2
n + 1 = 0
n = -1
â´ n = -2 , -1
n^2 + 2n + n + 2 = 0
(n^2 + 2n) + (n + 2) = 0
n(n + 2) + 1(n + 2) = 0
(n + 2)(n + 1) = 0
n + 2 = 0
n = -2
n + 1 = 0
n = -1
â´ n = -2 , -1
2009-04-21 10:50 pm
The easy way is factoring.
(n + 2)(n + 1) = 0
n = -1, -2
n = [-3 +/- sqrt(3^3 - 4*2)]/2
n = [-3 +/- sqrt(9 - 8)]/2
n = [-3 +/- sqrt 1]/2
n = [-3 +/- 1]/2
n = (-3 - 1)/2 = -4/2 = -2
n = (-3 + 1)/2 = -2/2 = -1
(n + 2)(n + 1) = 0
n = -1, -2
n = [-3 +/- sqrt(3^3 - 4*2)]/2
n = [-3 +/- sqrt(9 - 8)]/2
n = [-3 +/- sqrt 1]/2
n = [-3 +/- 1]/2
n = (-3 - 1)/2 = -4/2 = -2
n = (-3 + 1)/2 = -2/2 = -1
2009-04-21 10:49 pm
0=(n+2)(n+1)
n+2=0 n+1=0
n= -2 n= -1
n+2=0 n+1=0
n= -2 n= -1
2009-04-21 10:49 pm
No. that is not the quadratic formula. Make it clear what you are asking.
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