Question

What are the solutions to the equation 2(x-3)^2=54 ?

Answer 1

Answer:

      x =(6-√108)/2=3-3√ 3 = -2.196

 x =(6+√108)/2=3+3√ 3 = 8.196

Step-by-step explanation:

Step  1  :

Equation at the end of step  1  :

2 • (x - 3)2 - 54 = 0

Step  2  :

 2.1    Evaluate :  (x-3)2   =  x2-6x+9 

Step  3  :

Pulling out like terms :

 3.1     Pull out like factors :

   2x2 - 12x - 36  =   2 • (x2 - 6x - 18) 

Adding  9  has completed the left hand side into a perfect square :

   x2-6x+9  =

   (x-3) • (x-3)  =

  (x-3)2  (x-3)1 =

   x-3

Now, applying the Square Root Principle to  Eq. #4.3.1  we get:

   x-3 = √ 27

Add  3  to both sides to obtain:

   x = 3 + √ 27

Since a square root has two values, one positive and the other negative

   x2 - 6x - 18 = 0

   has two solutions:

  x = 3 + √ 27

   or

  x = 3 - √ 27

Solve Quadratic Equation using the Quadratic Formula

 4.4     Solving    x2-6x-18 = 0 by the Quadratic Formula .

 According to the Quadratic Formula,  x  , the solution for   Ax2+Bx+C  = 0  , where  A, B  and  C  are numbers, often called coefficients, is given by :

                                     

            - B  ±  √ B2-4AC

  x =   ————————

                      2A

  In our case,  A   =     1

                      B   =    -6

                      C   =  -18

Accordingly,  B2  -  4AC   =

                     36 - (-72) =

                     108

Applying the quadratic formula :

               6 ± √ 108

   x  =    —————

                    2

Can  √ 108 be simplified ?

Yes!   The prime factorization of  108   is

   2•2•3•3•3 

To be able to remove something from under the radical, there have to be  2  instances of it (because we are taking a square i.e. second root).

√ 108   =  √ 2•2•3•3•3   =2•3•√ 3   =

                ±  6 • √ 3

  √ 3   , rounded to 4 decimal digits, is   1.7321

 So now we are looking at:

           x  =  ( 6 ± 6 •  1.732 ) / 2

Two real solutions:

 x =(6+√108)/2=3+3√ 3 = 8.196

or:

 x =(6-√108)/2=3-3√ 3 = -2.196