Completing the square?

16x^2 - 12x - 1 = 0
To complete the square:

1. move the constant (term with no x co-efficient) to the other side.
2. make the x^2 term coefficient equal to 1.
3. take half of the x term coefficient , square it, and add to both sides of equation.
4. you can now factor into squared form and continue to solve.

16x^2 - 12x = 1
x^2 - (12/16)x = 1/16
x^2 - (3/4)x + 9/64 = 1/16 + 9/64
(x - 3/8)^2 = (4 + 9)/64
(x - 3/8)^2 = 13/64
x - 3/8 = sqrt(13/64)
x = sqrt(13/64) + 3/8
x = (1/8)[sqrt(13) + 3]

16x²-12x-1=0
(4x-3/2)²-13/4=0
(4x-3/2)²=13/4
4x-3/2=±√(13/4)
4x=3/2±√13/2
4x=(3±√13)/2
x=(3±√13)/8
x1=(3+√13)/8, x2=(3-√13)/8

You should start this one by squaring out things like:

(4x+1)^2
(4x-1)^2
(4x+3)^2

just to get an idea as to what a perfect square might look like.

Then you can ask yourself, "What value for h would make (4x+h)^2 start out with a 16x^2 -12x?" Once you have that, the value you need to add to 16x^2 -12x to make it a perfect square should be obvious.

By the way, 16 x^2 - 12x + (9/4) should do the trick.
That's equal to (4x - 3/2)^2.
You'll have to add 13/4 to each side of your equation to get it to work.

After you understand what's going on with completing the square, you're going to realize that the b^2-4ac part of the quadratic formula must be zero when you have a perfect square. You can use this to find a formula for c in terms of a and b. Since you're going to use completing the square to derive the quadratic formula, it's really bad form mathematically to rely on it to complete the square... but it works fine.

16x^2 - 12x - 1 = 0
4(4x^2 - 3x) - 1 = 0
4[(2x - (3/4))^2 - (9/16)] - 1 = 0
4[(2x - (3/4))^2] - 9/4 - 1 = 0
4(2x - (3/4))^2 = 13/4
(2x - (3/4))^2 = 13/16
2x - (3/4) = sqrt(13/16)
2x = sqrt(13/16) + 3/4
2x = [3 + sqrt(13)] / 4
x = [3 + sqrt(13)] / 8

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Answer
x = (3±√13)/8

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..... 16x²-12x-1=0
....... 4²x²-12x = 1
The middle term of a perfect square (a+b)² is 2ab.
Letting a=4, then 2(4)b=-12 → b= -3/2 and b²=9/4

4²x²-12x+9/4 = 1+9/4
. . . (4x-3/2)² = 13/4 ...now take the square root
. . . . . 4x-3/2 = ±(√13)/2 ...now multiply by 2
. . . . . . 8x-3 = ±√13
. . . . . . . . 8x = 3±√13
. . . . . . . . . x = (3±√13)/8

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16x^2-12x-1=0=>
16(x^2-(3/4)x+(3/8)^2-(3/8)^2)-1=0=>
16(x^2-3/8)^2-9/4-1=0=>
16(x^2-3/8)^2-(sqrt(13/4))^2=0=>
[4(x-3/8)+sqrt(13)/2][4(x-3/8)-
sqrt(13)/2]=0=>
x=[3-sqrt(3)]/8 or
x=[3+sqrt(13)]/8

16x^2 - 12x - 1 = 0
x = [-b ±√(b^2 - 4ac)]/2a

a = 16
b = -12
c = -1

x = [12 ±√(144 + 64)]/32
x = [12 ±√208]/32
x = [12 ±√(4^2 * 13)]/32
x = [12 ±4√13]/32
x = [3 ±√13]/8

∴ x = [3 ±√13]/8

Equation: 16x^2-12x-1=0
16(x^2-12/16 x)-1
16(x^2-12/16 x+(12/32)^2)-1-16(12/32)^2
16(x-12/32)^2-1-144/64
16(x-12/32)^2-208/64
16(x-12/32)^2-208/64=0
16(x-3/8)^2 -13/4 =0
16(x-3/8)^2 =13/4
(x-3/8)^2 = 13/64
(x-3/8) = +/-sqrt(13/64)
x= 3/8 +/- sqrt(13)/8

16x² - 12x - 1 = 0

Add 1 to both sides.
16x² - 12x - 1 + 1 = 0 + 1
16x² - 12x = 1

Factor
16[x² - (3 / 4)x] = 1

Add placeholders.
16[x² - (3 / 4)x + ___] = 1 + 16(___)

Notice that the second blank is multiplied by 16 to account for the 16 you factored out.

Take the coefficient of the x term: (-3 / 4)
Divide it by 2: (-3 / 4) / 2 = (-3 / 8)
Square it: (-3 / 8)² = 9 / 64

Add (9 / 64) to both blanks.
16[x² - (3 / 4)x + (9 / 64)] = 1 + 16(9 / 64)

x² - (3 / 4)x + (9 / 64) is the expanded form of a perfect square binomial.

Remember that (a - b)² = a² - 2ab + b². Apply this to what you have.
16[x² - (3 / 4)x + (9 / 64)] = 1 + 16(9 / 64)
16[x - (3 / 8)]² = 1 + 16(9 / 64)

Simplify the rest.
16[x - (3 / 8)]² = 1 + (9 / 4)
16[x - (3 / 8)]² = 1(4 / 4) + (9 / 4)
16[x - (3 / 8)]² = (4 / 4) + (9 / 4)
16[x - (3 / 8)]² = (4 + 9) / 4
16[x - (3 / 8)]² = (13 / 4) <== completed square

Divide both sides by 16.
16[x - (3 / 8)]² / 16 = (13 / 4) / 16
[x - (3 / 8)]² = (13 / 64)

Take the square root of both sides.
√[x - (3 / 8)]² = √(13 / 64)
x - (3 / 8) = ± (√13 / 8)

Add (3 / 8) to both sides.
x - (3 / 8) + (3 / 8) = (3 / 8) ± (√13 / 8)
x = (3 / 8) ± (√13 / 8)
x = (3 ± √13) / 8

ANSWER: x = (3 ± √13) / 8

CHECK USING QUADRATIC FORMULA:

Given: ax² + bx + c = 0
Quadratic Formula: x = [-b ± √(b² - 4ac)] / 2a

Given: 16x² - 12x + -1 = 0
Means: a = 16, b = -12, c = -1

x = [-b ± √(b² - 4ac)] / 2a
x = [-(-12) ± √((-12)² - 4(16)(-1))] / 2(16)
x = [12 ± √(144 + (64)] / 32
x = [12 ± √(208)] / 32
x = [12 ± √208] / 32
x = [12 ± √(16 * 13)] / 32
x = (12 ± 4√13) / 32
x = (3 ± √13) / 8
TRUE

16 [ x ² - (12/16) x ] - 1 = 0

[ x ² - (3/4) x ] = 1/16

[ x ² - (3/4) x + 9/64 ] = 1/16 + 9/64

[ x - (3/8) ] ² = 13/64

x - (3/8) = ± (1/8) √13

x = (3/8) ± (1/8) √13

x = (1/8) [ 3 ± √13 ]