Answer: The correct answer is: " 1 " .
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→ " x² + 2x + <u> 1 </u> = (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 + <u> 1 </u> = (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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