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

Question

Looking to solve for x.

Answer 1

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

Answer 2

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

:)

Answer 3

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