Question

How do you solve `x ^ { 2} - 8x + 5`?

Answer 1

Complete the square to find:

`x^2-8x+5 = (x-4-sqrt(11))(x-4+sqrt(11))`

which has zeros:

`x = 4+sqrt(11)" "` and `" "x = 4-sqrt(11)`

Explanation:

Given:

`x^2-8x+5`

There is nothing to solve here. This is not an equation to solve, neither does the question ask for factorisation or zeros.

I guess you would like to find the zeros, i.e. the solutions of the equation:

`x^2-8x+5 = 0`

Note that `x^2-8x+5` is in the standard form `ax^2+bx+c` with `a=1`, `b=-8` and `c=5`.

This has discriminant `Delta` given by the formula:

`Delta = b^2-4ac`

`color(white)(Delta) = (color(blue)(-8))^2-4(color(blue)(1))(color(blue)(5))`

`color(white)(Delta) = 64-20`

`color(white)(Delta) = 44`

Since `Delta > 0` we can tell that this quadratic has two distinct real zeros, but since `Delta` is not a perfect square, those zeros are irrational.

We can factor the quadratic by completing the square and using the difference of squares identity:

`A^2-B^2 = (A-B)(A+B)`

with `A=(x-4)` and `B=sqrt(11)` as follows:

`x^2-8x+5 = x^2-2(4)x+16-11`

`color(white)(x^2-8x+5) = (x-4)^2-(sqrt(11))^2`

`color(white)(x^2-8x+5) = ((x-4)-sqrt(11))((x-4)+sqrt(11))`

`color(white)(x^2-8x+5) = (x-4-sqrt(11))(x-4+sqrt(11))`

Hence zeros:

`x = 4+-sqrt(11)`