# Completing the square when the x² value is more than 1

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#1
How do i complete the square when the x² value is more than one? For example if it was 3x² + 2x - 1

I had a really bad teacher last year and I have mocks in like 3 weeks and I still don't get it so any help would be greatly appreciated. Thanks
0
2 months ago
#2
(Original post by thrivingfrog)
How do i complete the square when the x² value is more than one? For example if it was 3x² + 2x - 1

I had a really bad teacher last year and I have mocks in like 3 weeks and I still don't get it so any help would be greatly appreciated. Thanks
Factorise the co-effecient of x^2 so that it becomes 1

so in this case it would be 3[x^2+2/3x-1/3]
1
#3
(Original post by the_pharaoh)
Factorise the co-effecient of x^2 so that it becomes 1

so in this case it would be 3[x^2+2/3x-1/3]
Ty!
0
2 months ago
#4
This used to get me so much glad I wasn't the only one
0
2 months ago
#5
(Original post by the_pharaoh)
Factorise the co-effecient of x^2 so that it becomes 1

so in this case it would be 3[x^2+2/3x-1/3]
I personally prefer the method where you do this:
3[x^2+2/3x]-1
Because it has less fractions to deal with, but they'll both get you the same answer in the end
2
#6
(Original post by englishhopeful98)
This used to get me so much glad I wasn't the only one
0
2 months ago
#7
(Original post by thrivingfrog)
0
#8
How would I solve the equation once I've completed the square?
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#9
(Original post by Roses.Are.Red)
Yes, thankyou
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2 months ago
#10
(Original post by thrivingfrog)
How would I solve the equation once I've completed the square?
put it equal to zero
0
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