Looking to solve for x.

Tutors, sign in to answer this question.

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

Tom D. | Very patient Math Expert who likes to teachVery patient Math Expert who likes to te...

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

:)

Already have an account? Log in

By signing up, I agree to Wyzant’s terms of use and privacy policy.

Or

To present the tutors that are the best fit for you, we’ll need your ZIP code.

Your Facebook email address is associated with a Wyzant tutor account. Please use a different email address to create a new student account.

Good news! It looks like you already have an account registered with the email address **you provided**.

It looks like this is your first time here. Welcome!

To present the tutors that are the best fit for you, we’ll need your ZIP code.

Please try again, our system had a problem processing your request.

## Comments