更新1:
OK, so Yeah, QUADRATIC EQUATION!!!!
Solving Equalities, can you help?
2008-12-26 12:42 am
OK, (2x^2) = 3x - 2. Do you subtract the x squared and use the quadratic equation? Or am I thinking of something else? I forgot how to isolate to just get an x. I mean you could get x = 3 but to double check it is wrong 18 doesn't equal 7.
回答 (8)
2008-12-26 12:59 am
✔ 最佳答案
take everything to one side and use the quadratic formula...roots are complex
2008-12-26 8:31 pm
2x² = 3x - 2
2x² - 3x = - 2
x² - 3/2x = - 1
x² - 3/4x = - 1 + (3/4)²
x² - 3/4x = - 16/16 + 9/16
(x - 3/4)² = - 7/16
x - 3/4 = 7i/4
x = (7i + 3)/4
x = (3 - 7i)/4
Answer: x = (7i - 3)/4, (3 - 7i)/4
2x² - 3x = - 2
x² - 3/2x = - 1
x² - 3/4x = - 1 + (3/4)²
x² - 3/4x = - 16/16 + 9/16
(x - 3/4)² = - 7/16
x - 3/4 = 7i/4
x = (7i + 3)/4
x = (3 - 7i)/4
Answer: x = (7i - 3)/4, (3 - 7i)/4
2008-12-28 12:37 am
2 x ² - 3 x + 2 = 0
x = [ 3 ± â (9 - 16) ] / 4
x = [ 3 ± â (- 7) ] / 4
x = [ 3 ± i â7 ] / 4
x = [ 3 ± â (9 - 16) ] / 4
x = [ 3 ± â (- 7) ] / 4
x = [ 3 ± i â7 ] / 4
2008-12-26 10:11 am
2x^2=3x-2
2x^2-3x+2=0
a=2 b=-3 c=2
-(-3)+/- sqrt -3^2-4(2)(2)/4
3+/- sqrt 9-16/4
3+/- sqrt -7/4
3+/- i sqrt 7/4
2x^2-3x+2=0
a=2 b=-3 c=2
-(-3)+/- sqrt -3^2-4(2)(2)/4
3+/- sqrt 9-16/4
3+/- sqrt -7/4
3+/- i sqrt 7/4
2008-12-26 9:09 am
2x² = 3x - 2
Transpose everything to the left
2x² - 3x + 2 = 0
Factoring. (I hope you already know how to do that)
It is not factorable so use the quadratic formula
x = {3 ± â[(-3)² - 4(2)(2)]}/2(2)
x = {3 ± â-7}/4
x = (3 ± iâ7)/4
x = (3 - iâ7)/4,
or x = (3 + iâ7)/4
I hate this. A nonfactorable quadratic equation with complex roots at that.
Transpose everything to the left
2x² - 3x + 2 = 0
Factoring. (I hope you already know how to do that)
It is not factorable so use the quadratic formula
x = {3 ± â[(-3)² - 4(2)(2)]}/2(2)
x = {3 ± â-7}/4
x = (3 ± iâ7)/4
x = (3 - iâ7)/4,
or x = (3 + iâ7)/4
I hate this. A nonfactorable quadratic equation with complex roots at that.
2008-12-26 9:00 am
2x^2 = 3x - 2
subtract 3x from both sides
2x^2 - 3x = -2
add 2 on both sides
2x^2 - 3x + 2 = 0
use quadratic eqn formula
x = [ 3 ± â(9 -16) ] /4
x = [ 3 ± â-7 ] /4
x = (1/4)[ 3 ± iâ7 ]
subtract 3x from both sides
2x^2 - 3x = -2
add 2 on both sides
2x^2 - 3x + 2 = 0
use quadratic eqn formula
x = [ 3 ± â(9 -16) ] /4
x = [ 3 ± â-7 ] /4
x = (1/4)[ 3 ± iâ7 ]
2008-12-26 8:58 am
2x^2 = 3x - 2
2x^2 - 3x + 2 = 0
x = [-b 屉(b^2 - 4ac)]/2a (<= quadratic formula)
a = 2
b = -3
c = 2
x = [3 屉(9 - 16)]/4
x = [3 屉(-7)]/4
x = [3 ±iâ7]/4
â´ x = [3 ±iâ7]/4
2x^2 - 3x + 2 = 0
x = [-b 屉(b^2 - 4ac)]/2a (<= quadratic formula)
a = 2
b = -3
c = 2
x = [3 屉(9 - 16)]/4
x = [3 屉(-7)]/4
x = [3 ±iâ7]/4
â´ x = [3 ±iâ7]/4
2008-12-26 8:53 am
2x^2 = 3X-2 substract and get 2x^2 - 3x - 2 = 0. basically u want to set to zero. now FOIL (X+?)(X+?) = 0
solve the question marks and thats how u do these
solve the question marks and thats how u do these
參考: took math all the way to calc
收錄日期: 2021-05-01 11:44:25
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20081225164240AAV2oRC