How do you complete the square?????

Question

x^2+12x+___


x^2-6x+___


x^2-6x=16


2x^2-3x+1=0

Answer 1

x^2 + 12x + ?? = (x+6)^2 (you need the 6 to get the 12x in the middle) = x^2 + 12x + 36

x^2 - 6x + ?? = (x-3)^2 = x^2 - 6x + 9

The next looks a lot like this one, but you'll have to borrow from the right side of the equal's signL
x^2 - 6x = (x-3)^2 - 9 = 16 or (x-3)^2 = 25
which means that (x-3) = 5 or -5, so x = 8 or -2

For the fourth, divide everything by 2 before trying!
x^2 - 3/2x + 1/2 = 0
(x-3/4)^2 - (3/4)^2 + 1/2 = 0
(x-3/4)^2 = (3/4)^2 - 1/2

Answer 2

for the first two equations, use this method...

let's say this is the form.... 1x^2 + Bx + C... we are looking for C...

take the coefficient B... divide it by 2... and square that term... that will be your C...

1. 12/2 = 6... 6^2 = 36

that makes your equation... x^2 + 12x + 36... if we factor this, we get... (x + 6)^2

2. -6/2 = -3... (-3)^2 = 9

that makes your equation... x^2 -6x + 9... if we factor this, we get... (x - 3)^2

for the third one, you do the same step for finding C... we find C and it is equal to the same as second... 9

x^2 - 6x + 9 = 16 + 9
(x - 3)^2 = 25

take square root of both sides...
x - 3 = + or - 5
x - 3 = 5 --> x = 8
x - 3 = -5 --> x = -2

the fourth one is a little different... there is just one more step... since the coefficient of x^2 is 2, we will divide the whole equation by 2...

we get x^2 -(3/2)x = -(1/2)

we will follow the same steps for C... [(-3/2)/2] = (-3/4)
C = (-3/4)^2 = 9/16

add it to the eq...

x^2 -(3/2)x + 9/16 = -1/2 + 9/16
x^2 -(3/2)x + 9/16 = 1/16

if we factor the left hand side, we get...
[x - (3/4)]^2 = 1/16

take square root of both sides...
x - (3/4) = + or - 1/4
x - (3/4) = 1/4 --> x = 4/4 = 1
x - (3/4) = -1/4 --> x = 2/4 = 1/2

Answer 3

Hi,

When your x^2 term has a 1 as its coefficient like your first 3 problems, all you have to do is divide the x term coefficient by 2 and square that number. That is what you would have to add to complete the square. So, in your first one 12/2 = 6__ 6^2=36, so you'd add 36 to complete the square. In your second one, -6/2 = -3 and (-3)^2 = 9, so you'd add 9 to complete the square.
For your third one, you'd find -6/2 = -3 and (-3)^2 = 9, so you'd add 9 to complete the square. This would add 9 to BOTH sides of the equation, so it would become x^2 - 6x + 9 = 16 +9. This could be simplified and solved, but you didn't ask for that.

For your last problem before you can complete the square you have to get rid of the 2 in front of the x^2 term by dividing every term by 2. Your equation would now be x^2 - (3/2)x + 1/2 = 0. Subtract 1/2 to move it to the other side, x^2 - (3/2)x = -1/2
Now take -3/2 divided by 2, which is -3/4. Then square it (-3/4)^2 = 9/16 and add 9/16 to both sides.

Answer 4

To complete the square you need to do 3 steps. But first make sure the the coefficient of x^2 is equal to one, if it isn't then divide out the coefficient. After that put the x^2 & bx term on one side of the equation and the constant on the other side.
I will be referring to a, b, and c; from the standard form of a quadratic equation ax^2+bx=c=0. So when you get to step one you are at this point x^2 +bx=c
Step 1. mulitiply 'b' by one half (or divide by 2)
Step 2. take the answer from step 1 and square it.
Step 3. add the answer from step 2 to both sides of the equation.

And that's it.
So let's work out #3 x^2-6x=16
x^2 and bx are one one side of the equation.
So b = -6. And -6(1/2) = -3. -3 squared is 9. So, you add 9 to both sides
x^2-6x+9=16+9
Now a short cut, the left side is now a perfect square trinomial. And it will always factor to be x & whatever half of 'b' is. So in this case we get.
(x-3)(x-3)=25 [adding 16+9]
Now we can rewrite the left side
(x-3)^2=25
To 'undo' the square we take the square root of both sides of the euqation
sqrt{(x-3)^2}=sqrt{25}
The squares and square roots cancel out leaving us
x-3 = sqrt{25}
the right side is a perfect square
x - 3 = 5
REMEMBER WHEN YOU TAKE THE SQUARE ROOT OF AN EQUATION YOU CREATE AN EXTRA ANSWER...ALWAYS HAVE A + and -....so
x - 3 = +/-5
add 3 to both sides to solve for x
x = 3 +/- 5
so x = 3 + 5 = 8 or
x = 3 - 5 = -2

For number 1 & 2 just do the 3 steps, they aren't equations.
Number 4, you need to divide everything by 2 first.

Answer 5

x^2 + 12x + 36 = (x+6)^2.

x^2 - 6x + 9 = (x-3)^2.

x^2 - 6x + 9 = 16 + 9 = 25, so (x-3)^2 = 25, and x-3 = +5 or -5, so x = 8 or -2.

2x^2 - 3x + 1 = 0 so x^2 - (3/2)x + (1/2) = 0
Complete square: x^2 - (3/2)x + (1/2 + 1/16 [=9/16]) = 1/16
(x-3/4)^2 = 1/16, so x-3/4 = +1/4 or -1/4, and x = 1 or 1/2.

Answer 6

1)x^2+12x+___
Divide 12 by 2 and then square it:
12/2=6
6^2=36
x^2+12x+36

2)x^2-6x+___
Divide -6 by 2 and then square it:
-6/2=-3
(-3)^2=9
x^2-6x+9

3)x^2-6x=16
-6/2=-3
(-3)^2=9
x^2-6x+9=16+9
(x-3)^2=25
√(x-3)^2=±√25
x-3=±5
x=±5+3
x=-2 and 8

Answer 7

(b/2)^2

x^2+12x+(12/2)^2
x^2+12x+36
(x+6)^2

x^2-6x+ (-6/2)^2
x^2-6x+9
(x-3)^2

x^2-6x=16
x^2-6x+(6/2)^2=16+(6/2)^2
x^2-6x+9=16+9
(x-3)^2= 25
x-3= 5 or -5
x= 8 and -2

2x^2-3x+1=0
x^2-3/2x+1/2
x^2-3/2x= -1/2
x^2-3/2x+(3/2/2)^2= -1/2+(3/2/2)^2
x^2-3/2x+9/16= -1/2+9/16
(x-3/4)^2= 1/16
x-3/4= sqr(1/16)
x= 3/4+sqr(1/16) or 3/4-sqr(1/16)

Answer 8

u mean the blank spaces?