4n^2 - 4n - 35 = 0
6m^2 + 19m + 10 = 0
I need to find the roots of each equation. Can you just tell me the numbers from the two parantheses? for example, the answer would be (3m+2)(m+5). i need something like that
Thanks in advance!
2007-08-25 04:29:59 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Question 1
(2n - 7) (2n + 5) = 0
n = 7 / 2 , n = - 5 / 2
Question 2
(3m + 2) (2m + 5) = 0
m = - 2 / 3 , m = - 5 / 2
2007-08-25 08:48:58 · answer #1 · answered by Como 7 · 2⤊ 0⤋
4n^2 - 4n - 35 = 0
(2n + 5)(2n - 7) = 0
6m^2 + 19m + 10 = 0
(2m + 5)(3m + 2) = 0
.
2007-08-25 11:41:24 · answer #2 · answered by Robert L 7 · 0⤊ 0⤋
There are some strategies to find the roots of the equations: trial and error (factoring), completing the square or using quadratic formula. You may use any of these method to find the roots.
2007-08-25 11:38:57 · answer #3 · answered by rochelle 2 · 0⤊ 0⤋
a)(2n -7)(2n + 5) = 0
n = 7/2 or -5/2
b)(3m + 2)(2m + 5) = 0
m= -2/3 or m = -5/2
2007-08-25 11:39:31 · answer #4 · answered by timemccormick 3 · 0⤊ 0⤋
4n^2 - 4n - 35 = 0
compare it an^2+bn+c=0
here a=4, b=-4, c=-35
its roots are
[n-{-b+â(b^2-4ac)}/2a][n-{-b-â(b^2-4ac)}/2a]=0
[n-{4+â(16+560)}/8][n-{4-â(16+560)}/8]=0
[n-{4+24}/8][n-{4-24}/8]=0
(n-7/2)(n+5/2)
similarly solve other one.
2007-08-25 11:42:24 · answer #5 · answered by Jain 4 · 0⤊ 0⤋
Hi. Remember that FOIL is your friend. You will need to learn the factoring technique and apply it. Just learn it once.
2007-08-25 11:36:25 · answer #6 · answered by Cirric 7 · 0⤊ 0⤋