How do I solve this
7i/3+4i
and this problem too
6x^2+13x=5
Thanks
2006-09-11 05:15:24 · 6 answers · asked by CPSweetie 3 in Education & Reference ➔ Homework Help
no it's not the first one is 7i divided by 3+4i....
I as in the imaginary number
2006-09-11 05:20:13 · update #1
1.) 7i/3 + 4i is incomplete, but you can simplify it by adding fractions:
7i/3 + 12i/3 = 19i/3
2.) 6x^2+13x=5
Step 1: Set the equation equal to 0.
6x^2 + 13x - 5 = 0
Step 2: Use quadratic equation:
x = (-b +/- (b^2-4ac)^1/2)/2a
x = (-13 +/- (169 + 120)^1/2)/12
x = -13/12 +/- 17/12
x = 4/12 , -30/12
x = 1/3, -2 1/2 (solution)
check:
6x^2 + 13x - 5 = 0
6 (1/3)^2 + 13 (1/3) - 5 = 0
6/9 + 13/3 - 5 = 0
0 = 0 (check)
6x^2 + 13x - 5 = 0
6 (-2.5)^2 + 13 (-2.5) - 5 = 0
37.5 - 32.5 - 5 = 0
0 = 0 (check)
2006-09-11 05:23:18 · answer #1 · answered by ³√carthagebrujah 6 · 1⤊ 0⤋
For the second one...
set it equal to zero...6x^2 +13x -5=0
and solve for x using FOIL or quadratic formula
The first one is incomplete (it's not equated to anything) and needs clarification...is it (7i/3)+4i or 7i/(3+4i)?
2006-09-11 05:27:11 · answer #2 · answered by czimme3 4 · 0⤊ 0⤋
The first question is incomplete but try to rationlise the denominator i.e . multiply denominater and nu. by ( 3 - 4i) and simplify.
2006-09-11 05:26:26 · answer #3 · answered by macline k 2 · 0⤊ 0⤋
The first one is incomplete.
2006-09-11 05:17:36 · answer #4 · answered by happytraveler 4 · 0⤊ 0⤋
first one, multiply by the conjegate.
second one, quadratic formula.
2006-09-11 05:21:21 · answer #5 · answered by sexy azn 2 · 0⤊ 0⤋
=Study and answer it
2006-09-11 05:17:28 · answer #6 · answered by Anonymous · 0⤊ 0⤋