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-1
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avatar+851 

When the expression $-2x^2-20x-53$ is written in the form $a(x+d)^2+e$, where $a$, $d$, and $e$ are constants, then what is the sum $a+d+e$?

Lightning  Jul 21, 2018
edited by Lightning  Jul 21, 2018
 #1
avatar+2 
0

ans is 

sag121  Jul 21, 2018
 #2
avatar+92826 
+1

When the expression $-2x^2-20x-53$ is written in the form $a(x+d)^2+e$, where $a$, $d$, and $e$ are constants, then what is the sum $a+d+e$?

 

-2x^2 - 20x - 53        we need to complete the square on x...factor out the "-2"

 

-2 (x^2  + 10x  + 53/2)      

 

Take 1/2 of 10  = 5....square it  = 25....add and subtract it inside the parentheses

 

-2 (x^2  +10x + 25  + 53/2  - 25)      factor the first three terms and simplify the rest

 

-2 [ ( x + 5)^2   + 3/2)      distribute the "-2"  back across the terms inside the parentheses

 

-2(x + 5)^2  - 3

 

So  a = -2, d  = 5  and e  = -3     and their  sum is  0

 

 

cool cool cool

CPhill  Jul 21, 2018

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