Find the values of a, b, and c for which the quadratic equation ax^2+bx+c=0 has these numbers as solutions:
13i, -13i
2006-10-18 20:12:21 · 8 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
(x+i13)(x-i13) =
x^2 -i13x +i13x +169
a=1
b=0
c=169
2006-10-18 20:18:04 · answer #1 · answered by Helmut 7 · 2⤊ 0⤋
ax^2 + bx = c = 0
as solution = 13i, -13i
it means that
(x + 13i)(x - 13i) = 0
=> x² - 13².i² = 0
But i² = -1
Thus x² + 169 = 0
So, a = 1, b = 0, c = 169
2006-10-19 04:09:23 · answer #2 · answered by nanduri p 2 · 0⤊ 0⤋
roots of the equation are 13i and -13i
so sum of the roots is 0
product of the roots is 169
so the equation will be of form,
x^2 - (sum of the roots)*x + product of the roots=0
so x^2 + 169 = 0
a = 1
b = 0
c = 169
2006-10-19 04:21:07 · answer #3 · answered by ksj_goblin 3 · 0⤊ 0⤋
b must be zero ..... (b/2 is zero in a + bi form)
with b = 0 .... we'll solve for c w/ a=1
discriminate must be 26i (26i / 2 yields +-13i)
-0+-(0-(4*1*169)^1/2 / (2*1)
equation is x^2 +169 = 0
hope this helps
2006-10-19 03:21:58 · answer #4 · answered by Brian D 5 · 0⤊ 0⤋
The equation would be (0.5)x^2+84.5=0. The x term is 0.
2006-10-19 03:21:33 · answer #5 · answered by Patrick W 1 · 0⤊ 0⤋
(x+i13)(x-i13) =
x^2 -i13x +i13x +169
a=1,b=0,c=169
2006-10-19 03:51:57 · answer #6 · answered by Anonymous · 0⤊ 0⤋
a=1
b=0
c=169
2006-10-19 03:22:48 · answer #7 · answered by Anonymous · 0⤊ 1⤋
Those are vector quantities, I believe in this quesiton they represent the x-intercepts! I have no idea how to get those numbers though, sorry.
2006-10-19 03:19:19 · answer #8 · answered by Anonymous · 0⤊ 2⤋