Solution - Quadratic equations

Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

                     68*b-b^2-(256)=0 

Step by step solution :

Step  1  :

Step  2  :

Pulling out like terms :

 2.1     Pull out like factors :

   -b² + 68b - 256  =   -1 • (b² - 68b + 256) 

Trying to factor by splitting the middle term

 2.2     Factoring  b² - 68b + 256 

The first term is,  b²  its coefficient is  1 .
The middle term is,  -68b  its coefficient is  -68 .
The last term, "the constant", is  +256 

Step-1 : Multiply the coefficient of the first term by the constant   1 • 256 = 256 

Step-2 : Find two factors of  256  whose sum equals the coefficient of the middle term, which is   -68 .

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -64  and  -4 
                     b² - 64b - 4b - 256

Step-4 : Add up the first 2 terms, pulling out like factors :
                    b • (b-64)
              Add up the last 2 terms, pulling out common factors :
                    4 • (b-64)
Step-5 : Add up the four terms of step 4 :
                    (b-4)  •  (b-64)
             Which is the desired factorization

Equation at the end of step  2  :

  (4 - b) • (b - 64)  = 0 

Step  3  :

Theory - Roots of a product :

 3.1    A product of several terms equals zero. 

 When a product of two or more terms equals zero, then at least one of the terms must be zero. 

 We shall now solve each term = 0 separately 

 In other words, we are going to solve as many equations as there are terms in the product 

 Any solution of term = 0 solves product = 0 as well.

Solving a Single Variable Equation :

 3.2      Solve  :    -b+4 = 0 

 Subtract  4  from both sides of the equation : 
                      -b = -4
Multiply both sides of the equation by (-1) :  b = 4

Solving a Single Variable Equation :

 3.3      Solve  :    b-64 = 0 

 Add  64  to both sides of the equation : 
                      b = 64

Supplement : Solving Quadratic Equation Directly

Solving    b2-68b+256  = 0   directly 

Earlier we factored this polynomial by splitting the middle term. let us now solve the equation by Completing The Square and by using the Quadratic Formula

Parabola, Finding the Vertex :

 4.1      Find the Vertex of   y = b²-68b+256

Parabolas have a highest or a lowest point called the Vertex .   Our parabola opens up and accordingly has a lowest point (AKA absolute minimum) .   We know this even before plotting  "y"  because the coefficient of the first term, 1 , is positive (greater than zero). 

 Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two  x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions. 

 Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex. 

 For any parabola,Ab²+Bb+C,the  b -coordinate of the vertex is given by  -B/(2A) . In our case the  b  coordinate is  34.0000  

 Plugging into the parabola formula  34.0000  for  b  we can calculate the  y -coordinate : 
  y = 1.0 * 34.00 * 34.00 - 68.0 * 34.00 + 256.0
or   y = -900.000

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = b²-68b+256
Axis of Symmetry (dashed)  {b}={34.00} 
Vertex at  {b,y} = {34.00,-900.00} 
 b -Intercepts (Roots) :
Root 1 at  {b,y} = { 4.00, 0.00} 
Root 2 at  {b,y} = {64.00, 0.00} 

Solve Quadratic Equation by Completing The Square

 4.2     Solving   b²-68b+256 = 0 by Completing The Square .

 Subtract  256  from both side of the equation :
   b²-68b = -256

Now the clever bit: Take the coefficient of  b , which is  68 , divide by two, giving  34 , and finally square it giving  1156 

Add  1156  to both sides of the equation :
  On the right hand side we have :
   -256  +  1156    or,  (-256/1)+(1156/1) 
  The common denominator of the two fractions is  1   Adding  (-256/1)+(1156/1)  gives  900/1 
  So adding to both sides we finally get :
   b²-68b+1156 = 900

Adding  1156  has completed the left hand side into a perfect square :
   b²-68b+1156  =
   (b-34) • (b-34)  =
  (b-34)²
Things which are equal to the same thing are also equal to one another. Since
   b²-68b+1156 = 900 and
   b²-68b+1156 = (b-34)²
then, according to the law of transitivity,
   (b-34)² = 900

We'll refer to this Equation as  Eq. #4.2.1  

The Square Root Principle says that When two things are equal, their square roots are equal.

Note that the square root of
   (b-34)²   is
   (b-34)^2/2 =
  (b-34)¹ =
   b-34

Now, applying the Square Root Principle to  Eq. #4.2.1  we get:
   b-34 = √ 900

Add  34  to both sides to obtain:
   b = 34 + √ 900

Since a square root has two values, one positive and the other negative
   b² - 68b + 256 = 0
   has two solutions:
  b = 34 + √ 900
   or
  b = 34 - √ 900

Solve Quadratic Equation using the Quadratic Formula

 4.3     Solving    b²-68b+256 = 0 by the Quadratic Formula .

 According to the Quadratic Formula,  b  , the solution for   Ab²+Bb+C  = 0  , where  A, B  and  C  are numbers, often called coefficients, is given by :
                                     
            - B  ±  √ B²-4AC
  b =   ————————
                      2A

  In our case,  A   =     1
                      B   =   -68
                      C   =  256

Accordingly,  B²  -  4AC   =
                     4624 - 1024 =
                     3600

Applying the quadratic formula :

               68 ± √ 3600
   b  =    ——————
                      2

Can  √ 3600 be simplified ?

Yes!   The prime factorization of  3600   is
   2•2•2•2•3•3•5•5 
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).

√ 3600   =  √ 2•2•2•2•3•3•5•5   =2•2•3•5•√ 1   =
                ±  60 • √ 1   =
                ±  60

So now we are looking at:
           b  =  ( 68 ± 60) / 2

Two real solutions:

b =(68+√3600)/2=34+30= 64.000

or:

b =(68-√3600)/2=34-30= 4.000

Two solutions were found :

1.  b = 64
2.  b = 4