Question

How do you solve `x^2-5x-6=0` by completing the square?

Answer 1

`x=-1" and "x=6`

Please do not change this solution. It is a reference solution.

Explanation:

`color(blue)("End objective")`

Given the standard form `y=ax^2+bx+c`

We end up with form `y=a(x+ b/(2a))^2+c+k`

The problem is that changing the original equation to this form introduces an error. So `k` is a necessary correction. The value of which is determined by:

`a(b/(2a))^2+k=0`

`color(magenta)("For this question "a=1)`
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`color(blue)("Completing the square")`

Given:`" "y=0=x^2-5x-6`.......................Equation(1)

`color(brown)("Step 1")`

Write as: `(x^2-5x)-6+k larr" At this stage "k" has no value"`

'...................................................................................................
`color(brown)("Step 2")`
Move the power from `x^2` to outside the brackets

`(x-5x )^(color(magenta)(2))-6+k`

`color(purple)(" Now "k" starts to take on values that change at each step")`

'...................................................................................................
`color(brown)("Step 3")`
Discard the `x` from `5x` inside the brackets

`(x-5)^2-6+k`

'...................................................................................................
`color(brown)("Step 4")`

Halve the 5 inside the brackets

`(x-5/2)^2-6+k`

'...................................................................................................
`color(brown)("Step 5")`

Determine the value of `k`
`(xcolor(magenta)(-5/2))^(color(magenta)(2))-6+color(magenta)(k)`......................Equation(2)

Let `(color(magenta)(-5/2))^(color(magenta)(2))+ color(magenta)(k)=0" "rarr" Note that this is the bit:"`
`color(white)("222222222222222222222222") "a(color(magenta)(b/(2a)))^(color(magenta)(2))+color(magenta)(k)=0`

`color(white)("ddddddddddddddddddddd")`where in this case `a=1`

`=>25/4+k=0" " =>" " k= -25/4`

So by substitution equation(2) becomes:

`y=0=(x-5/2)^2-6-25/4`

`color(blue)(y=0=(x-5/2)^2-49/4)color(white)(.)......Equation(2_a)`
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`color(brown)("Step 6 - Determine the vertex")`

Using `Eqn(2_a)`
Vertex`->(x,y)=[(-1)xx(-5/2)color(white)("ddd"),color(white)("d")-49/4]`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Solving for "y=0)`

Add `49/4` to both sides

`(x-5/2)^2=49/4`

Square root both sides

`x-5/2=+-sqrt(49/4) = +-7/2`

Add `5/2` to both sides

`x=5/2+-7/2`

`x=-1" and "x=6`

Answer 2

As Tony B. states the answer is `x=6` or `x=-1`. The explanation here uses a different method to complete the square.

Explanation:

In this method a different approach is used to complete the square. The idea is to show more clearly how the coefficient `5` gets "halved".

Start with the expression

`x^2-5x=x(x-5)`

These expression is the same as the first two terms of the quadratic expression originally given. We render this as a difference of squares:

`x(x-5)=u^2-v^2=(u+v)(u-v)`

And then we match the two sets of factors:

`x=u+v`

`x-5=u-v`

Now add these two equations together and note that in eliminating `v` we make a coefficient of `2` on `u`. This is how we end up "halving the `5`:

`2x-5=2u` so `u=(x-5/2)`

Then take the difference of the original two equations to get `v` which is just a constant, since `u` and `x` cancel at the same time:

`5=2v` so `v=5/2`; again half of `5` appears.

This works because the original quadratic expression had a coefficient of `1` on `x^2`. If the original quadratic expression has a different coefficient, like `2` in `2x^2-10x-12=0`, we would have divided by that coefficient on `x^2` at the beginnjng to get (in this case) `x^2-5x-6=0` with a coefficient of `1` on `x^2`, and then we could proceed as above.

Then we have:

`x^2-5x=u^2-v^2=(x+5/2)^2-({25}/4)`

And the original equation becomes:

`x^2-5x-6=0`

`(x-5/2)^2-({25}/4)-6=0`

`(x-5/2)^2=({49}/4)`

So we take both possible square roots noting that `{49}/4` happens to be a perfect square:

`(x-5/2)=+(7/2)` so `x=6` or

`(x-5/2)=-(7/2)` so `x=-1`

If we did not have a perfect square the final answer would contain square roots.