Question

What is the missing constant term in the perfect square that starts with x^2+2x

Answer 1

Answer:  The correct answer is:  " 1 " .

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                               →   "  x²  +  2x  +    1     =   (x + 1)²  " .

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Step-by-step explanation:

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Let us assume that the question asks us to solve for the "missing constant term in the following equation:  

      →   " x² + 2x + b = 0 " ;  

      →  in which:  " b " is the "missing constant term" for which we shall solve.

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The form of an equation in the perfect square would be:

                →  (x + b) ² =  x² + 2bx + b²  ;

                →  In our case, "b" ; refer to the "missing constant term" for which we shall solve.

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      →   " x² + 2x + b = 0 " ;  

       Note that the term in the equation with the highest degree (highest exponent) is:  

            →  " x² " ;  with an "implied coefficient" of: " 1 " (one) ;

      →   {since "any value" , multiplied by " 1 " , results in that same initial value.}.

      →   Since the term with the highest degree has a "co-efficient" of " 1 " ;

  we can solve the problem;   i.e. "Solve for "b" ;  accordingly:

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      →   " x² + 2x + b = 0 " ;

Subtract "b" from each side of the equation:

      →   " x² + 2x + b - b = 0 - b " ;

      →  to get:

      →   x² + 2x  =  - b

Now we want to complete x² + 2x into a perfect square.

To do so, we take the:  "2" (from the:  "+2x" );  

     →  and we divide that value {in our case, "2"};  by "2" ;

to get:  "[2/2]" ;  and then we "square" that value;

     →  to get:  " [2/2]² " .

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Now, we add this "squared value" to:  " x² + 2x " ;  as follows:

     →  " x² + 2x + [2/2]² " ;  and simply:  " [2/2]² = [1]² = 1 ."

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     x² + 2x + (2/2)² = x² + 2x + 1 ;

                              =  (x + 1)² ;

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Now:  " x² + 2x = - b " ;

 We add "(2/2)² " ;  to each side of the equation;

 →  In our case,  " [2/2]² = [1]² = 1 " ;

 →  As such, we add:  " 1 " ;  to each side of the equation:

              →   x² + 2x + (2/2)² =  - b + (2/2)² ;

              →  Rewrite;  substituting " 1 " [for:  " (2/2)² "] :

              →   x² + 2x + 1  =  1 - b ;                                          

              →   x² + 2x + 1   = 1 - b ;

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And assume "b" would equal "1" ;

 since assuming the question refers to the equation:

     "x²  +  2x  ±   b = 0 " ;  solve for "b" ;  

And:   "b = 1 " ;

Then:  " x² + 2x + 1 = ?  1 - b  ??

       →  then:   " 1 - b = 0 "  ;  Solve for "b" ;

       →  Add "b" to each side of the equation:

                      " 1 - b + b = 0 + b " ;

       →  to get:  " 1 = b "  ;   ↔ " b =  1 "  ;  Yes!

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Also, to check our work:

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Remember, from above:

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" The form of an equation in the perfect square would be:

                →  (x + b)²  =  x²  +  2bx + b² " ;  _____________________________________________________

   →  Let us substitute "1" for all values of "b" :

                  →   "  (x + 1) ²  =   x² + 2*(b)*(1)  +   1²  "  ;

                  →   "  (x + 1)²   =   x²  +  (2*1*1)   +  (1*1) "  ;

                  →   "  (x + 1)²   = ?  x²  +  2  +  1 "  ?? ;  Yes!

                  →  However, let us check for sure!

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→   Expand:  " (x + 1)²  " ;  

→  " (x + 1)² = (x + 1)(x + 1) " ;

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   →  " (x + 1)(x + 1) " ;  

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Note the following property of multiplication:

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 →   " (a + b)(c + d)  =  ac  +  ad  +  bc  +  bd " ;

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As such:

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         →  " (x + 1)(x + 1) " ;

              =  (x*x) + (1x) + (1x) + (1*1) ;

              =  x²  +  1x + 1x + 1 ;

        →  Combine the "like terms" :

              + 1x + 1x = + 2x ;  

And rewrite:

              =   x²  +  2x + 1 .

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"  (x + 1)²   = ?  x²  +  2  +  1 "  ?? ;  Yes!

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    →   So:  The answer is:  " 1 " .

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   →  " x² + 2x +  1    =  (x + 1)²  " .

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Hope this answer helped!

    Best wishes to you in your academic endeavors

            — and within the "Brainly" community!

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Answer 2

Answer:

  1

Step-by-step explanation:

The constant term in a perfect square trinomial with leading coefficient 1 is the square of half the coefficient of the linear term.

  (2/2)² = 1

The missing constant term is 1.