Question

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

Answer 1

`color(blue)(x=5)` or `color(blue)(x=1)`

Explanation:

Remember in squaring a general binomial
`color(white)("XXX")(a+b)^2=a^2+2ab+b^2`

Given: `x^2-6x+5=0`

If `x^2` and `-6x` are the first two terms of the expansion of a squared binomial:
`color(white)("XXX")a=1`, and
`color(white)("XXX")b=-3`

So the third term must be `b^2=(-3)^2=9`

To complete the square we add a `9` to the first two terms (but to keep everything balanced we will need to subtract it again:
`color(white)("XXX")x^2-6xcolor(red)(+9)+5color(red)(-9)=0`

Now we can write the first 3 terms as a squared binomial:
`color(white)("XXX")(x-3)^2+5-9=0`

Simplifying `(5-9)=-4` and adding `4` to both sides gives
`color(white)("XXX")(x-3)^2=4`

Taking the square root of both sides
`color(white)("XXX")x-3=+-2`

Then add `3` to both sides
`color(white)("XXX")x=3+-2`

`rArr x=5 or x=1`

Answer 2

`x = 5, " or " x =1`

Explanation:

It seems strange to want to use completing the square, because this quadratic trinomial factorises to give `(x-5)(x-1) =0`

But let's proceed... Completing the square is based on the fact that
the square of a binomial gives a standard answer

`(x -8)^2 = x^2 -color(blue)(16)x +color(magenta)64`

There is ALWAYS a relationship between the `color(blue)16 and color(magenta)64`

This is: `color(blue)(16 div 2)" and then squared gives "color(magenta)64 " "8^2 =64`

In `x^2 -6x + 5 = 0 " the relationship is not correct"`

This means that 5 is not the correct constant, move it to the RHS

`x^2 -color(blue)(6)x color(magenta)(+9)= -5color(magenta)(+9)" add the correct constant to both sides"`

`(x-3)^2 = 4 " now the LHS is a perfect square"`

`x-3 = +-sqrt4" square root both sides"`

`x = 2+3, " or " x = -2+3`

`x = 5, " or " x =1`

Answer 3

Another example of method. Have a look at:

https://socratic.org/s/awchNPwZ