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I came up against this problem

 

If we express 2x^2 + 6x + 11 in the form a(x - h)^2 + k, then what is h?

 

Any help would be greatly appreciated!

 Oct 23, 2017
 #1
avatar
+1

Complete the square:
2 x^2 + 6 x + 11

Group terms containing x, leaving a placeholder constant:
(2 x^2 + 6 x + __) + 11 - __

(2 x^2 + 6 x + __) = 2 (x^2 + 3 x^2 + __):
2 (x^2 + 3 x + __) + 11 - __

Take one half of the coefficient of x and square it. Then add and subtract, multiplying by the factored constant 2 on the outside.
Insert (3/2)^2 = 9/4 on the inside and subtract (2×9)/4 = 9/2 on the outside:
2 (x^2 + 3 x + 9/4) + 11 - 9/2

11 - 9/2 = 13/2:
2 (x^2 + 3 x + 9/4) + 13/2

x^2 + 3 x + 9/4 = (x + 3/2)^2:
2(x + 3/2)^2 + 13/2         and h=3/2

 Oct 23, 2017
 #2
avatar+272 
0

Hey thanks for the response but that didn't work.  Was there a typo in your answer or is my hw wrong (thats possible, its happened before)

 Oct 23, 2017
 #3
avatar+98196 
+2

 2x^2 + 6x + 11   factor out the 2

 

2 [ x ^2 + 3x  + 11/2]   

Take 1/2  of 3  =  3/2....square it  = 9/4    ....add it and subtract it inside the parentheses

 

2 [ x^2 + 3x + 9/4  +  11/2 - 9/4 ]  factor  the first three terms and simplify

 

2 [ ( x + 3/2)^2  +  22/4 - 9/4]

 

2 [ (x + 3/2)^2  +  13/4 ]      distribute the 2

 

2 ( x + 3/2)^2  +  13/2

 

The guest made a slight error.......we can write this as

 

2 [ x - (-3/2) ]^2  + 13/2

 

So....h  = -3/2

 

 

cool cool cool

 Oct 23, 2017

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