Find 3 consecutive odd numbers where the product of the smaller two numbers is 46 less than the square of the largest number.

 The numbers are:  " 5, 7, and 9 " .


Explanation:

Let the "3 consecutive odd numbers" be represented by:

 "x" ;  

"x + 2 " ; 

"x + 4 " ; 

→ Solve for "x" ; and then solve for "(x + 2)" and "(x + 4)" .


The product of the smaller two numbers, is "46 less than" (the square of) the largest number: 

→  " x* (x + 2)   =  (x + 4)² − 46 " ; 

Note the "distributive property" of multiplication:

  a(b + c)  =  ab + ac ; 

  a(b − c)  =  ab − ac .


Start with simplifying the "left-hand side" of the equation:


 " x(x + 2)" = (x * x) + (x * 2 ) = (x² + 2x) ;

Then simplify the "right-hand side" of the equation:


 " (x + 4)² − 46 " ;

 →  (x + 4)² =  (x + 4) (x + 4) ; 

Note:  " (a + b)(c + d) = ac + ad + bc + bd " .

 →  (x + 4)² =  (x + 4) (x + 4) = (x *x) + (x *4) + (4 *x) + (4 * 4) ; 

                                             = x² + 4x + 4x + 16 ; 
      
                                             = x² + 8x + 16; 

           →  Bring down the " − 46 " ; 

→  x² + 8x + 16 − 46 ;

=    x² + 8x + 16 − 46 ;

=    x² + 8x − 30 ; 

Now, rewrite the entire equation:


→  x² + 2x = x² + 8x − 30 ;

Subtract "x² " & "2x" from EACH SIDE of the equation:

→  x² + 2x − x² − 2x = x² + 8x − 30 − x² − 2x ; 

to get:

→  0  = 6x − 30 ; 

→  6x − 30 = 0 ; 

Add "30" to each side of the equation:

→  6x − 30 + 30 = 0 + 30 ; 

→  6x = 30 ; 

Divide each side of the equation by "6" ; 
       to isolate "x" on each side of the equation; & to solve for "x" ; 

→  6x / 6 = 30 / 6 ;

to get;

→  x = 5 ;

→ "(x + 2)" = "(5 + 2)" = 7 ;

→ "(x + 4)" = "(5 + 4)" =  9 ;
 
 The numbers are:  " 5, 7, and 9 " .

Element a has no reactivity because it has eight valence electrons and cannot gain anymore electron and it wouldn't lose any either