# How would you solve equation? [x_2_2] [2_x_2]=0 [2_2_x]

*print*Print*list*Cite

### 1 Answer

You need to evaluate determinant of given matrix to form the equation that you need to solve such that:

`[[x,2,2],[2,x,2],[2,2,x]]` = `x^3 + 8 + 8 - 4x - 4x - 4x`

You need to solve the equation `[[x,2,2],[2,x,2],[2,2,x]] = ` 0 such that:

`x^3 + 8 + 8 - 4x - 4x - 4x = 0`

You need to collect like terms such that:

`x^3 + 16 - 12x = ` 0

Rearranging the terms yields:

`x^3 - 12x + 16 = 0`

You need to verify if one of divisors of 16 may be a root for the equation. Selecting x=2 yields:

`2^3 - 12*2 + 16 = 8 - 24 + 16 = 0`

Notice that x=2 is the solution to equation `x^3 - 12x + 16 = 0` , hence you may write the factorized form of equation such that:

`x^3 - 12x + 16 = (x-2)(ax^2 + bx + c)`

You need to open the brackets to the right side such that:

`x^3 - 12x + 16 = ax^3 + bx^2 + cx - 2ax^2- 2bx - 2c`

Equating coefficients of like powers both sides yields:

`a=1`

`b-2a=-12 =gt b-2=-12 =gt b=-10`

`-2c=16 =gt c = -8`

Hence, evaluating the factor `ax^2 + bx + c` yields `x^2- 10x - 8` .

You need to find two more roots to equation `x^3 - 12x + 16 = 0` , hence, you need to solve equation `x^2 - 10x - 8 = 0` such that:

`x_(2,3) = (10+-sqrt(100 + 32))/2`

`x_2 = 5+2sqrt33; x_3 = 5-2sqrt33`

**Hence, evaluating solutions to equation `x^3 - 12x + 16 = 0` yields `x_1=2; x_2 = 5+2sqrt33; x_3 = 5-2sqrt33` .**