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

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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

quadratic equation form ax^2+bx+c= 0

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.**

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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