Question

How do you solve `x^2 + 10x + 13 = 0` by completing the square?

Answer 1

`" "x~~ -8.464" and "-1.536`

Explanation:

Given: `y=x^2+10x+13=0`

This process introduces an error that has to be compensated for. To do this I introduce a corrective as yet unknown value represented by `k`. The value of `k` may be determined after all the other changes have taken place

Compare to `y=ax^2+bx+c`

This then written as a first step as:

`y=a(x^2+b/ax)+c` in your case `a=1`

`color(blue)("Step 1")`

`y=(x^2+10x)+13+k_1 larr" At this point "k_1=0`

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`color(blue)("Step 2")`

Move the square from `x^2` outside the brackets. This begins to introduce errors.

`y=(x+10x)^2+13+k_2`

`color(brown)(ul("Just for Evan"))`
`color(brown)(k_2" is whatever is needed to turn "y=(x+10x)^2+13`
`color(brown)("back to "y=x^2+10x+13`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Step 3")`

Remove the `x` from the `10x`

`y=(x+10)^2+13+k_3`

`color(brown)(ul("Just for Evan"))`
`color(brown)(k_3" is whatever is needed to turn "y=(x+10)^2+13)`
`color(brown)("back to "y=x^2+10x+13`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Step 4")`

Halve the 10

`y=(x+5)^2+13+k_4`

`color(brown)(ul("Just for Evan"))`
`color(brown)(k_4" is whatever is needed to turn "y=(x+5)^2+13)`
`color(brown)("back to "y=x^2+10x+13`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
NOW WE DETERMINE THE VALUE OF `k`

Just for demonstration lets multiply out the brackets and then compare what we have to the original equation.

`y=x^2+10x color(red)(+25) +13+k_4 larr" New equation"`
`y=x^2+10x color(white)("dd.d")+13 color(white)("ddd.")larr" Original equation"`

For the new equation to work we must have `25+k_4=0 =>color(purple)(k_4=-25)`
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`color(blue)("Step 5 : Substitute for "k_4)`

`color(green)(y=(x+5)^2+13color(purple)(+k_4) color(white)("ddd") -> color(white)("ddd") y= (x+5)^2+13color(purple)(-25) )`

`color(green)(color(white)("dddddddddddddddddddd") ->color(white)("dd")y=(x+5)^2-12`
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`color(blue)("Now we solve for "x)`

Set `" " y=0= (x+5)^2-12`

Add 12 to both sides

`12=(x+5)^2`

Square root both sides

`+-sqrt(12)=x+5`

Subtract 5 from both sides

`x=-5+-2sqrt(3)`

Answer 2

`x = -5 pm2sqrt(3)`

Explanation:

Start with `x^2+10x+13=0`

Subtract 13 from both sides:

`x^2+10x=-13`

Take 10, divide by 2, to get 5. Square 5 to get 25. Add 25 to both sides:

`x^2+10x +25=-13+25`

`x^2+10x +25=12`

The left side is now a perfect square:

`(x+5)^2=12`

Now solve by taking the square root of both sides:

`x+5 = pmsqrt(12)`

Subtract 5 from both sides:

`x = -5 pmsqrt(12)`

Simplify the radical if you'd like:

`x = -5 pm2sqrt(3)`