+267 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!
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
+267 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)
+129852 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
