# Given f (x) = 6x^2 + x and g(x) = 3− 2x , find all values of x such that f (x) = g(x)?

lemjay | Certified Educator

calendarEducator since 2012

starTop subjects are Math and Science

To determine the values of x that will make f(x) and g(x) equal, the steps are:

f(x)    =   g(x)        Set f(x) equal to g(x).

6x^2 + x =  3 - 2x       Move " 3-2x" to the left side to have in

6x^2 + x - 3 + 2x = 0

6x^2  + 3x - 3 = 0       Divide the both sides by 3 to simplify.

2x^2  + 3x - 1 = 0       Use quadratic formula to solve for x.

-b +/- sqrt(b^2 - 4ac)          -3 +/- sqrt[3^2 - 4(2)(1)]

x=  ------------------------- =   ------------------------------

2a                                         2(2)

-3 +/- sqrt( 9 - 8)       -3 +/- sqrt(1)       -3 +/- 1

x = ---------------------- =  ---------------  =   -------

4                               4                      4

x1 =  (-3 + 1) / 4             and      x2 = (-3 - 1) / 4

x1 =  -2 / 4                                 x2 = -4 / 4

x1 = -1/2                                    x2 = -1

Answer: The values of x that make the two functions equal are -1/2 and -1.

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sciencesolve | Certified Educator

calendarEducator since 2011

starTop subjects are Math, Science, and Business

You need to solve the equation `6x^2+x = 3-2x ` to find the values of x that verify both `f(x)`  and `g(x)`  such that:

`6x^2+x = 3-2x =gt 6x^2 + x - 3 + 2x = 0`

You need to accumulate like terms such that:

`6x^2 + 3x - 3 = 0`

You need to divide by 3 such that:

You need to use quadratic formula such that:

`x_(1,2) = (-b+-sqrt(b^2-4ac))/(2a)`

You need to identify the values of coefficients a,b,c, hence ,comparing `ax^2 + bx + c = 0`  to `2x^2 + x - 1 = 0` , yields `a=2,b=1,c=-1` .

`2x^2 + x - 1 = 0`

You need to substitute 2 for a, 1 for b and -1 for c in quadratic formula such that:

`x_(1,2) = (-1+-sqrt(1+8))/4`

`x_(1,2) = (-1+-sqrt9)/4`

`x_(1,2) = (-1+-3)/4`

`x_1 = -1/2 ; x_2 = -1`

Hence, evaluating the values of x that make `f(x)`  equal to `g(x)`  yields `x_1 = -1/2 ; x_2 = -1` .

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