Solution - Factoring binomials using the difference of squares

Rearrange:

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

                     2*22/7*r^2-(77)=0 

Step by step solution :

Step  1  :

            22
 Simplify   ——
            7 

Equation at the end of step  1  :

        22            
  ((2 • ——) • r2) -  77  = 0 
        7             

Step  2  :

Equation at the end of step  2  :

  44r2    
  ———— -  77  = 0 
   7      

Step  3  :

Rewriting the whole as an Equivalent Fraction :

 3.1   Subtracting a whole from a fraction

Rewrite the whole as a fraction using  7  as the denominator :

          77     77 • 7
    77 =  ——  =  ——————
          1        7   

Equivalent fraction : The fraction thus generated looks different but has the same value as the whole

Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator

Adding fractions that have a common denominator :

 3.2       Adding up the two equivalent fractions
Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

 44r2 - (77 • 7)     44r2 - 539
 ———————————————  =  ——————————
        7                7     

Step  4  :

Pulling out like terms :

 4.1     Pull out like factors :

   44r² - 539  =   11 • (4r² - 49) 

Trying to factor as a Difference of Squares :

 4.2      Factoring:  4r² - 49 

Theory : A difference of two perfect squares,  A² - B²  can be factored into  (A+B) • (A-B)

Proof :  (A+B) • (A-B) =
         A² - AB + BA - B² =
         A² - AB + AB - B² =
         A² - B²

Note :  AB = BA is the commutative property of multiplication.

Note :  - AB + AB equals zero and is therefore eliminated from the expression.

Check :  4  is the square of  2 
Check : 49 is the square of 7
Check :  r²  is the square of  r¹ 

Factorization is :       (2r + 7)  •  (2r - 7) 

Equation at the end of step  4  :

  11 • (2r + 7) • (2r - 7)
  ————————————————————————  = 0 
             7            

Step  5  :

When a fraction equals zero :

 5.1    When a fraction equals zero ...

Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.

Now,to get rid of the denominator, Tiger multiplys both sides of the equation by the denominator.

Here's how:

  11•(2r+7)•(2r-7)
  ———————————————— • 7 = 0 • 7
         7        

Now, on the left hand side, the  7  cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :
   11  •  (2r+7)  •  (2r-7)  = 0

Theory - Roots of a product :

 5.2    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.

Equations which are never true :

 5.3      Solve :    11   =  0

This equation has no solution.
A a non-zero constant never equals zero.

Solving a Single Variable Equation :

 5.4      Solve  :    2r+7 = 0 

 Subtract  7  from both sides of the equation : 
                      2r = -7
Divide both sides of the equation by 2:
                     r = -7/2 = -3.500

Solving a Single Variable Equation :

 5.5      Solve  :    2r-7 = 0 

 Add  7  to both sides of the equation : 
                      2r = 7
Divide both sides of the equation by 2:
                     r = 7/2 = 3.500

Two solutions were found :

1.  r = 7/2 = 3.500
   
2.  r = -7/2 = -3.500