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When the expression \(3x^2-24x+55\) is written in the form \(a(x+d)^2+e\), where a,d , and e are constants, then what is the sum a+d+e?

mathtoo  Aug 27, 2018

Best Answer 

 #1
avatar+7336 
+3

3x2 - 24x + 55

                                                      Factor  3  out of the first two terms.

   =   3(x2 - 8x) + 55

                                                      Add  16  and subtract  16  inside the parenthesees.

   =   3(x2 - 8x + 16 - 16) + 55

                                                      Factor  x2 - 8x + 16  as a perfect square trinomial.

   =   3( (x - 4)2 - 16) + 55

                                                      Distribute the  3  to the terms in parenthesees.

   =   3(x - 4)2 - 48 + 55

                                                      Combine  -48  and  55  to get  7 .

   =   3(x - 4)2 + 7

 

   =   a(x + d)2 + e

 

So...

 

a + d + e   =   3 + -4 + 7   =   6

hectictar  Aug 27, 2018
 #1
avatar+7336 
+3
Best Answer

3x2 - 24x + 55

                                                      Factor  3  out of the first two terms.

   =   3(x2 - 8x) + 55

                                                      Add  16  and subtract  16  inside the parenthesees.

   =   3(x2 - 8x + 16 - 16) + 55

                                                      Factor  x2 - 8x + 16  as a perfect square trinomial.

   =   3( (x - 4)2 - 16) + 55

                                                      Distribute the  3  to the terms in parenthesees.

   =   3(x - 4)2 - 48 + 55

                                                      Combine  -48  and  55  to get  7 .

   =   3(x - 4)2 + 7

 

   =   a(x + d)2 + e

 

So...

 

a + d + e   =   3 + -4 + 7   =   6

hectictar  Aug 27, 2018
 #2
avatar+621 
+4

Correct! That's how you do it!

mathtoo  Aug 27, 2018

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