1.determine K if -3 is root of 2x^2+k x+9=0
2.solve for B. A= 3B/KB+D
I tried looking on various websites but they're confusing.
Please help and explain your steps! Thanks.
2007-11-14 06:25:13 · 5 answers · asked by imok 1 in Science & Mathematics ➔ Mathematics
1. if x= -3 2 x (-3)^2 -3k + 9 = 0
18 -3k + 9 = 0
3k = 27
k = 9
2. A = 3B/KB + D
A - D = 3/K the B's can be cancelled
K(A-D) = 3
if K is 9 again
9(A-D) = 3
A-D = 3/9
or 1/3
2007-11-14 06:37:38 · answer #1 · answered by Anonymous · 0⤊ 0⤋
If -3 is a root, then
2(-3)^2 +k(-3) +9 = 0
-3k = -27
k = 9
A= 3B/(KB+D) <-- You must mean this, else it makes no sense as the B's would cancel out.
A(KB+D)= 3B
AKB +AD = 3B
AD = 3B -AKB
AD = B(3-AK)
AD/(3-AK) = B
2007-11-14 06:38:41 · answer #2 · answered by ironduke8159 7 · 0⤊ 0⤋
A= 3B/KB+D
B can be any number as they cancel out. Are u sure u got the qn right?
2007-11-14 06:33:55 · answer #3 · answered by tj is cool 5 · 0⤊ 0⤋
1) square the whole equation first
9= 2x^2 +kx +9
move the 9 over
2x^2 +kx= 0
factorise
x(2x +k)=0
k= -2x
2007-11-14 06:37:47 · answer #4 · answered by Just me 5 · 0⤊ 1⤋
evn i tried
2007-11-14 06:35:22 · answer #5 · answered by poori 2 · 0⤊ 1⤋