Question

How do you solve `100x^2-800x+1500=0`?

Answer 1

`x = 5` and `x = 3`

Explanation:

First, divide both sides of the equation by `100` to make this simpler to work with:

`1/100(100x^2 - 800x + 1500) = 0/100`

`(100x^2)/100 - (800x)/100 + 1500/100 = 0`

`x^2 - 8x + 15 = 0`

Now we can play with multipliers of 15 (1x15, 3x5, 5x3, 15x1) to factor the quadratic equation:

`(x - 5)(x - 3) = 0`

We can now solve each term for `0`:

`x - 5 = 0`

`x - 5 + 5 = 0 + 5`

`x - 0 = 5`

`x = 5`

and

`x - 3 = 0`

`x - 3 + 3 = 0 + 3`

`x - 0 = 3`

`x = 3`

Answer 2

`color(green)(x=3)` or `color(green)(x=5)`

Explanation:

If
`color(white)("XXX")100x^2-800x+1500=0`
then (after dividing both sides by `100`)
`color(white)("XXX")x^2-8+15=0`

We would like to factor this in the form:
`color(white)("XXX")(x-a)(x-b)=0`
and since
`color(white)("XXX")(x-a)(x-b)=x^2+(-a-b)x+ab`
we are looking for values of `a` and `b`
such that
`color(white)("XXX")-a-b=-8 rarr a+b=8`
and
`color(white)("XXX")ab=15`

Checking factors of `15`, we quickly find the pair `3` and `5` that satisfy our requirement.

So we have
`color(white)("XXX")(x-3)(x-5)=0`

From which it follows that
#{: ("either ",(x-3)=0," or ",(x-5)=0), (,rarr x=3,,rarrx=5) :}#