Question

How do you solve `x^2-8x+5=0`?

Answer 1

We use a process called completing the square, which works for all quadratic equations. Here it gives `x=4\pm \sqrt{11}`.

Explanation:

Rather than memorizing a formula, you should understand how "completing the square" works.

Render the equation as:

`x^2-8x=-5`

Then add a constant, which we will call `b^2`:

`color(blue)(x^2-8x+b^2=b^2-5)`

Now compare the left side with the identity

`(a+b)^2=a^2+2ab+b^2`

We then have

`x^2-8x+b^2=(a+b)^2`

if

`a^2=x^2` so set `a=x`

`2ab=-8x` so `b=-4`

So,go back to the equation in blue and put in `b=-4`, then:

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

So `x-4=\pm \sqrt{11}` meaning

`x=4\pm \sqrt{11}`.

Answer 2

Root 1 `color(blue)( = 4+sqrt(11)`

Root 2 `color(blue)( = 4-sqrt(11)`

Explanation:

Given:

`color(red)(x^2-8x+5=0`

The standard form of a quadratic is `color(blue)(ax^2+bx+c=0`

So, we have `color(blue)(a=1, b=-8 and c=5.`

Solutions are given by `color(red)([-b+-sqrt(b^2-4ac)]/(2a)`

`rArr [-(-8)+-sqrt((-8)^2-4(1)(5)]]/(2*1)`

Using scientific calculator, we get,

`rArr [8+-sqrt(44)]/(2` [ Intermediate Result 1 ]

Observe that `sqrt(44)` can also be simplified as

`sqrt(4*11)`

`rArr sqrt(2^2*11)`

`rArr sqrt(2^2)*sqrt(11)`

`rArr 2sqrt(11)`

Now, we can write [ Intermediate Result 1 ] as

`color(blue)((8+-2sqrt(11))/2]`

We can rewrite the above as

`color(blue)[8/2+-(2sqrt(11))/2]`

`(4*2)/2+-(2sqrt(11))/2`

`(4*cancel(2))/cancel(2)+-(cancel(2)sqrt(11))/cancel(2)`

`rArr 4+-sqrt(11)`

Hence, we have the following two solutions:

Root 1 `color(blue)( = 4+sqrt(11)`

Root 2 `color(blue)( = 4-sqrt(11)`

Using a scientific calculator, we can simplify the above results.

Root 1 `color(green)(~~ 7.31662479`

Root 2`color(green)(~~ 0.68337521`

You can verify the solutions visually by examining the image of the graph below:

Hope it helps.