Question

How do you solve `y^2 + 10y + 35 = 0` by completing the square?

Answer 1

`y=-5+sqrt(-10)`
`y=-5-sqrt(-10)`

Explanation:

`y^2+10y+35=0`
or
`y^2+10y+25+10=0`
or
`y^2+2(y)5+5^2+10=0`
or
`(y+5)^2=-10`
or
`y+5=+-sqrt(-10)`
or
`y=-5+-sqrt(-10)`
or
`y=-5+sqrt(-10)`========Ans `1`
or
`y=-5-sqrt(-10)`========Ans `2`

Answer 2

`y=-5+-(sqrt(-10)) i`

Explanation:

This is a quadratic in `y` instead of in `x`. This means that the plot will be rotated `90^0` and be of shape `sub` or `sup`.

Set as `" "x=y^2+10y+35`

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

Write as:`" "x=(y^2+10y)+35+k`

The `k` will be needed to correct an error that this approach introduces.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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

Take the index (power) outside the brackets
`x=(y+10y)^2+35+k`

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

Discard the `y` from `10y`
`x=(y+10)^2+35+k`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Step 4")`
Halve the 10
`x=(y+5)^2+35+k`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Step 5")`
Now we need to correct for the error. The 5 in `(y+5)^2` produces `5^2` which is a value that is not in the original equation

So `5^2+k=0 => k=-25` giving:

`color(brown)(x=(y+5)^2+35+k)color(blue)(" "->" "x=(y+5)^2+35-25`

`color(green)(x=(y+5)^2+10) " "larr" Completed the square"`

This is also known as the Vertex Form Equation
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Solving for "(y+5)^2+10=0)`

`(y+5)^2=-10`

Square root both sides

`y+5=+-sqrt(-10)`

As soon as you see a root of a negative it means Complex Numbers are involved. In other words; the plot does not cross the y-axis in this case. If the graph was of general shape `uu" or "nn` it would not cross the x-axis. This graph is of shape `sub`.

`y=-5+-(sqrt(-10)) i`