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!

WhichWitchIsWhich
Oct 23, 2017

#1**+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**

Guest Oct 23, 2017

#2**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)

WhichWitchIsWhich
Oct 23, 2017

#3**+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

CPhill
Oct 23, 2017