COMPLETING THE SQUARE!???!!??!! help.?

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BoMember since: 10 February 2007

Total points: 133 (Level 1)

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Bo
S COMPLETEING THE SQUARE?? help, please?
CAN SOMONE PLEASE EXPLAIN HOW TO COMPLETE THE SQUARE ON:

3xsquare - x


and if possible:

2xsquares - 5x

Either the answer in the bk is wrong, or i am wrong.....

Answer 1

3x^2 - x = 3(x^2 - x/3) = 3( (x-1/6)^2 -1/36)

2x^2-5x = 2(x^2-5/2) = 2( (x-5/4)^2 -25/16)

Answer 2

Consider any second degree curve Y = ax^2 + bx + c.

This can be rewritten as Y = (dx+e)^2 + f and you can work out easily that

a=d^2
b = 2*d*e
c = e^2-f

The advantage of completing a square comes in the proof of the well known formula X =( -b+- SQR(b^2-4ac))/2*a

Here is the proof

aX^2 + bX + c = 0
Subtract c from both sides
aX^2 +bX = -c
Divide by a
x^2 + bX/a = -c/a
complete the square by adding b^2/4a^2
(x + b/2*a)^2 = -c/a +b^2/4a^2
Take the Square root - remembering you can have postive or negative values
x + b/2*A = +- SQR((b^2 - 4*a*c)/4a^2)
Move b/2A to RHS
x = -b/2*A +- SQR(b^2-4ac)/2A
Now put them over a common denominator
x = (-b +-SQR(b^2-4ac))/2a

Answer 3

3xsquare - x
= 3x^2 - x
= 3{x^2 - (1/3)x}
= 3{[x - (1/6)]^2 - (1/36)}
= 3[x-(1/6)]^2 - (1/12)

and

2xsquares - 5x
= 2x^2 - 5x
= 2{x^2 - (5/2)x}
= 2{[x-(5/4)]^2 - 25/16}
= 2[x-(5/4)]^2 - 25/8

Answer 4

3x² - x
3(x² - x / 3)
3.[(x² - x / 3 + 1/36) - 1/36]
3.[ (x - 1/6) ² - 1 / 36 ]
3.(x - 1/6)² - 1 / 12

2x² - 5x
2(x² - (5 / 2).x)
2.[(x² - (5/2).x + 25/16) - 25 / 16 ]
2.[(x - 5/4)² - 25/16]
2.(x - 5/4)² - 25 / 8

Answer 5

2xsquare
-3xsquares