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Completing The Square

How do I complete the square of this equation
A=-1/4(x^2-2x-35)
Original post by freddiek69
How do I complete the square of this equation
A=-1/4(x^2-2x-35)


This is what I did:

A = -1/4(x^2-2x-35)

Step 1 (multiply out the bracket):
A = -1/4x^2 - 1/2x - 35/4

Step 2 (write it in the form of a(x+b/2a)^2+d):
A = -1/4(x+1)^2 + d

Step 3 (expand the square):
(original equation = the square to be completed)
A = -1/4x^2 - 1/2x - 35/4 = -1/4(x^2 + 2x + 1) + d

Step 4 (cancel out the same terms (underlined)):
A = -1/4x^2 - 1/2x - 35/4 = -1/4x^2 -1/2x - 1/4 + d

Step 5 (rearrange and solve to get d):
-1/4 + d = -35/4
d = -35/4 + 1/4
d = -36/4
d = -9

Step 6 (substitute d into the completed square):
A = -1/4(x+1)^2 - 9

I hope I'm right :biggrin:
Reply 2
Avatar for freddiek69
freddiek69
OP
6
Original post by PredictableParadigmShift
This is what I did:

A = -1/4(x^2-2x-35)

Step 1 (multiply out the bracket):
A = -1/4x^2 - 1/2x - 35/4

Step 2 (write it in the form of a(x+b/2a)^2+d):
A = -1/4(x+1)^2 + d

Step 3 (expand the square):
(original equation = the square to be completed)
A = -1/4x^2 - 1/2x - 35/4 = -1/4(x^2 + 2x + 1) + d

Step 4 (cancel out the same terms (underlined)):
A = -1/4x^2 - 1/2x - 35/4 = -1/4x^2 -1/2x - 1/4 + d

Step 5 (rearrange and solve to get d):
-1/4 + d = -35/4
d = -35/4 + 1/4
d = -36/4
d = -9

Step 6 (substitute d into the completed square):
A = -1/4(x+1)^2 - 9

I hope I'm right :biggrin:


I'm just a bit confused about where the d has come from.
Original post by freddiek69
I'm just a bit confused about where the d has come from.


d is just used because it is the missing thing you have to solve in order to complete the square, could have been x or whatever.
Reply 4
Avatar for freddiek69
freddiek69
OP
6
Original post by Bill Nye
d is just used because it is the missing thing you have to solve in order to complete the square, could have been x or whatever.


Ahhh ok thanks for the help.
Original post by PredictableParadigmShift
....


Bloody hell... Bit overkill, no? Also you went wrong at the beginning when you chose the square form.

Original post by freddiek69
How do I complete the square of this equation
A=-1/4(x^2-2x-35)


You can complete the square on x2−2x−35x^2-2x-35 first to give (x−1)2−36(x-1)^2-36 then replace it in the expression to get −14[(x−1)2−36]-\frac{1}{4}[(x-1)^2-36] then distribute the -1/4 into it to get −14(x−1)2+9-\frac{1}{4}(x-1)^2+9
Original post by RDKGames
Bloody hell... Bit overkill, no? Also you went wrong at the beginning when you chose the square form.



You can complete the square on x2−2x−35x^2-2x-35 first to give (x−1)2−36(x-1)^2-36 then replace it in the expression to get −14[(x−1)2−36]-\frac{1}{4}[(x-1)^2-36] then distribute the -1/4 into it to get −14(x−1)2+9-\frac{1}{4}(x-1)^2+9


ohh nice

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