2006-07-26 14:37:51 · 11 answers · asked by needalgebrahelp 1 in Education & Reference ➔ Homework Help
I got it to 13x^2 + 38 x + 25=0 but I forget the formula to get the real imaginaries out of there.
(13x +25) (x+1)
It would equal out 13x^2+38x+ 25
I rock.
2006-07-26 14:42:09 · answer #1 · answered by AprilRocksIt 3 · 0⤊ 0⤋
6x= (19x+25)/(x+1)
6x(x+1)=19x+25
6x^2 + 6x - 19x = 25
6x^2 - 13x - 25 = 0
add to quadratic equation as a=6 , b= -13 , c = 25.
x = {-b +-[sq.r. (b^2 - 4ac)]}/2a
b^2 - 4ac < -1
therefore sq.r. (b^2 - 4ac) is a pure imaginary number
therefore x = pure imaginary number
2006-07-26 23:40:03 · answer #2 · answered by average joe 1 · 0⤊ 0⤋
6x = 19x+25 over x+1
6x * (x+1) = 19x+25
6x^2 + 6x = 19x +25
0 = -6x^2 +13x +25
Now use quadratic equation
a= -6
b=13
c=25
2006-07-26 21:46:01 · answer #3 · answered by math guru 4 · 0⤊ 0⤋
6x = 19x+25/(x+1)
Rearrange to get an equation that equals zero:
You will get something like: 6x^2 - 13x -25 = 0
You can solve for x using the quadratic formula:
x = [ -b +/- sqrt(b^2 - 4ac)]/2a
The answer should be x = 3.39 or -1.23
2006-07-26 22:03:04 · answer #4 · answered by Kris 2 · 0⤊ 0⤋
Of what I remember from high school, I got 13x=25 over x+1, then from there I'm not SURE but I think it's 1.923
2006-07-26 21:50:22 · answer #5 · answered by Anonymous · 0⤊ 0⤋
An imaginary solution let x=banana
2006-07-26 21:41:13 · answer #6 · answered by Anonymous · 0⤊ 0⤋
6 +or- i times the square root of 89 all over 6. WOW it's difficult to write that answer in text!
2006-07-26 23:18:37 · answer #7 · answered by MollyMAM 6 · 0⤊ 0⤋
6x^2-13x-25. now use the quadratic formula. i think.
2006-07-26 21:43:39 · answer #8 · answered by A.I. 3 · 0⤊ 0⤋
Too much effort and id rather not relive this year's alg. II class again.
2006-07-26 21:42:18 · answer #9 · answered by Fitz 2 · 0⤊ 0⤋
whats the matter can't figure it out yourself?
2006-07-26 21:40:55 · answer #10 · answered by kcracer1 5 · 0⤊ 0⤋