complete the square expressions?

Question

Can someone tell me how to complete the square on an expression that starts with an integer? For example:

6 - 3x - x^2

thanks!

Answer 1

-(x^2+2*3/2x+9/4-9/4)+6
-(x+3/2)^2+33/4

Answer 2

To COMPLETE THE SQUARE:

6 - 3x -x^2 = 0 Multiply throughout by -1

-6 +3x +x^2 = 0 Take the constant -6 to the other side

3x + x^2 = 6 Add the SQUARE of HALF the coefficient of x to BOTH sides

9/4 + 3x +x^2 = 6 + 9/4 The Left hand side is now a perfect square and may be written as:

(x + 3/2)^2 = 8.25 i.e. (6 + 9/4)

Square root of BOTH sides gives:

x + 3/2 = + or - Square root of 8.25

x = - 3/2 + or - Square root of 8.25

Answer 3

y = 6 - 3x - x^2
y = -x^2-3x+6 --- rearrange to have x^2, x and constant in order.
y = -(x^2 + 3x - 6) --- make the first term x^2, with sign or number.
y = -[(x + 3/2)^2 - 33/4]
y = -(x + 3/2)^2 + 33/4

Another example.
y = 2x^2 - 4x + 6
y = 2(x^2 - 2x + 3)
y = 2[(x - 1)^2 + 2]
y = 2(x-1)^2 + 4

Got it?

Answer 4

to solve such an expression take out the negative sign...
6-3x-x^2= -(x^2+3x-6)
then solve and multiply the negative sigh to one of the factors.
OK

Answer 5

for any quadratic equation you can solve it by using the quadritic formula:
for ax² + bx + c = 0

x = (-b +/- sqrt (b^2 - 4ac) ) / (2a).

Answer 6

you have to be a nerd to know that