Question

How do you solve `2x^2-12x=-14` by completing the square?

Answer 1

`" "x~~4.414" and "x~~1.586` to 3 decimal places

Explanation:

Write as `2x^2-12x+14=0`

`color(red)("Method for completing the square in detail")`

`color(blue)("Completing the square")`

What process we are about to do introduces a value that is not in the original equation. So we mathematically compensate for this by the inclusion of a correction value. This correction value would turn the introduced error into 0 if we were to carry out the addition.

Suppose we had `2z+56`. Then suppose we added 4 which is viewed as an error. However if we write `2z+56+4-4` we have corrected the error

Let `k` be the constant of correction

Write as `" "2(x^2-6x)+14+k=0`

Move the power of 2 to outside the bracket

`" "2(x-6x)^2+14+k=0`

divide the 6 from `6x` by 2

`" "2(x-6/2 x)^2+14+k=0`

Discard the `x` from `6/2`

`" "2(x-6/2)^2+14+k=0`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The error comes from the `-6/2` This is inside a bracket that is squared and then multiplied by the 2 outside the bracket

So the error is `2xx(-6/2)^2 = +36/2=+18`

This has to be turned into 0 by `k` so `" "k+18=0 =>k=-18`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`color(brown)(2(x-12/2)^2+14+k=0)color(green)(""->" "2(x-6/2)^2+14-18=0)`

`" "color(blue)(bar(underline(|" "2(x-3)^2-4=0" "|)))`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine x intercepts")`

Write as:`" "(x-3)^2=4/2`

Square root both sides

`" "x-3=+-sqrt(2)`

`" "x=+3+-sqrt(2)`

`color(blue)(" "x~~4.414" and "x~~1.586" to 3 decimal places")`

Answer 2

`x = sqrt2 + 3` OR `x = -sqrt2 + 3`
`x= 4.414` OR `x= 1.586` ( 3 dec places)

Explanation:

Completing the square is based on the consistency of the answers to the square of a binomial.

`(x - 3)^2 = x^2 - 6x + 9`
`(x - 5)^2 = x^2 - 10x + 25`
`(x + 6)^2 = x^2 + 12x + 36`
In all of the products above, `ax^2+ bx + c` we see the following:

`a = 1`
The first and last terms, `a and c` are perfect squares.'
There is a specific relationship between 'b' - the coefficient of the `x` term and 'c'. Half of b, squared equals c.

Knowing this, it is always possible to add in a missing value for `c` (ie complete the square) to make the square of a binomial, which can then be written as `(x+?)^2`

In `2x^2 -12x = -14`, -14 has already been moved to the right hand side. The equation must be divided by 2 to make `a` = 1

This gives:

`x^2 -6x = -7`

NOW the correct value of `c` can be added to BOTH sides of the equation to give the answer to the square of a binomial.

`x^2 - 6x + color(red)(9)` = `-7 + color(red)(9) rArr [9 = (-6÷2)^2]`
`(x - 3)^2 = 2 rArr` where -3 is from `-6÷2`
`x - 3 = +-sqrt2 rArr` take the square root of both sides

This gives 2 possible answers for `x`.

`x = sqrt2 + 3` OR `x = -sqrt2 + 3`