Find 4 values of x (x^2-6x+4) = 16

Question

Find four values 

Answer 1

Hi Nataly,

 

Here is how we do this question:

 

The degree of the fully distributed polynomial is 3: therefore there can only be 3 solutions. 

 

x(x²-6x+4) = 16

 

The first solution that jumps out is that x=0 that first x in front of the parentheses creates that solution. Now we analyze the remaining quadratic

 

x²-6x+4=16

          -16  -16                                                 Subtract 16 from each side

 

x²-6x-12=0

 

Now we have a quadratic that we can perhaps factor? Let's see. We are looking for factors of 12 that add up to 6. Note that since the sign of 12 is negative, the signs of the two constants in the two binomial factors will be opposite.

 

The factors of 12 are: 1 and 12; 2 and 6; 3 and 4. There is no way to combine those pairs to obtain the result of 6. Thus, this does not factor so we use the quadratic formula

 

a=1; b=-6; c=-12

 

x=(1/2a)*(-b +/- sqrt (b²-4ac))                              A bit hard to write here but I trust you are familiar

 

plugging in a, b, and c

 

x = (1/2(1))*(-(-6) ± sqrt((-6)²-4(1)(-12))

   =0.5*(6 ± sqrt(36+48))

 

Let's separate the two terms:

 

x = 0.5*6 ± 0.5 * sqrt(84)

   = 3 ± 0.5*sqrt(84)

    

 

We need to simplify that square root: 84=4*21.

 

x = 3 ± 0.5*sqrt(4*21) = 3 ± 0.5* sqrt (4)*sqrt (21) = 3 ± 0.5*2*sqrt(21) = 3 ± sqrt (21)

 

So, in summary, the 3 solutions to the original equation are

 

x = 0, 3+sqrt(21), 3-sqrt(21)