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3x^2=2(3x+1)

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3 Answers

Or we can complete the square;
i.e., force x^2 - 2hx + h^2 and
replace with (x - h)^2:
 
3x^2 = 2(3x + 1) = 6x + 2
3x^2 - 6x = 2
x^2 - 2(1)x = 2/3
 
h = 1; add h^2 to both sides:
x^2 - 2(1)x + 1 = 2/3 + 1 = 5/3
 
Replace perfect square trinomial with square:
(x - 1)^2 = 15/9
 
Take square root of both sides (sqrt(n^2) = |n|):
|x - 1| = sqrt(15)/3
 
Solve for x:
x - 1 = ±sqrt(15)/3
x = 1 ± sqrt(15)/3 exactly
1)Place in standard form aX^2+bX+c=0
3X^2 - 6X - 2 =0
 
2)Solution is the famous quadratic formula (memorize it)  X= [ -b ± sqrt(b^2-4ac) ]/2a
 
X= [-(-6) ± sqrt(36-4(3)(-2) ]/ (2*3)
X = [6 ± sqrt(60) ]/6
X = 1 ± 2sqrt(15)/6
X = 1 ± sqrt(5/3)   <---Exact
X = (2.291, -.291)  <---Approximately 
3x^2 = 2(3x+1)
3x^2 = 6x + 2
 
3x^2 - 6x -2 = 0
 
Its a quadratic equation
 
Standard quadaratic equation
 
ax^2 + bx + c = 0
 
Use standard solution formula
 
x = (-b +/- sqrt(b^2 - 4.a.c))/2a
 
a = 3, b= -6, c=-2
 
substitute
 
x = (6 +/- sqrt ( -6^2 - 4.3.(-6))/2.3
 
x= (6+/- sqrt (80))/6
x= (6 +/- 9)6
 
x= 15/6 , -1/2
:)

Comments

Has a couple mistakes that are easy to make