Question

I need to show how to find these solutions 1/4 and 3/4 by completing the square!

Answer 1

Answer:

See explanation, and ask for more details if unclear!

Step-by-step explanation:

The perfect square of this equation is $x^2-x+\frac{1}{4}$, since the square would be $(x-\frac{1}{2})^2$. 1/4=4/16, meaning that you can set up the equation in the following way:

$(x^2-x+\frac{1}{4})-\frac{1}{16}=0$

$(x-\frac{1}{2})^2=\frac{1}{16}$

Take the square root of both sides:

$x-\frac{1}{2}=\pm \frac{1}{4}$

Add 1/2 to both sides:

$x=\frac{1}{2}\pm \frac{1}{4}=\frac{3}{4}, \frac{1}{4}$. Hope this helps!

Answer 2

Answer:

see below

Step-by-step explanation:

x^2 - x + 3/16 = 0

Subtract 3/16 from each side

x^2 - x + 3/16 - 3/16 =  -3/16

x^2 -x = -3/16

Take the coefficient of x

-1

Divide by 2

-1/2   call this a

Square it

(-1/2)^2 = 1/4

Add it to each side

x^2 - x +1/4 =- 3/16+1/4

Changing to the "square"

( x +a)^2 =  - 3/16+1/4

getting a common denominator

( x -1/2) ^2 = -3/16+ 4/16

 ( x -1/2) ^2 =1/16

Take the square root of each side

sqrt(  ( x -1/2) ^2)  = ± sqrt(1/16)

x -1/2 = ± 1/4

Add 1/2 to each side

x -1/2 + 1/2 = 1/2 ± 1/4

x = 1/2 ± 1/4

Separate into 2 equations

x = 1/2 + 1/4                 x = 1/2 - 1/4

    2/4 +1/4                          2/4 -1/4

        3/4                                    1/4