Question

How do you find the roots, real and imaginary, of `y=(x-5)(-7x-2)` using the quadratic formula?

Answer 1

There are two Real roots `x=+5` and `x=-2/7`

Explanation:

The roots are the values of `x` for which `y=0`
These are easy to determine from the given factored form (see bottom)
but if you need to demonstrate the use of the quadratic formula:

First you need to convert the factored form `(x-5)(-7x-2)`
into standard form (by multiplying the factors together and arranging in standard sequence):
`color(white)("XXX")color(red)(""(-7))x^2+color(blue)(33)x+color(green)(10)`
then apply the quadratic formula
`color(white)("XXX")x= (-color(blue)b+-sqrt(color(blue)b^2-4color(red)acolor(green)c))/(2color(red)a)`

Substituting corresponding values for `color(red)a,color(blue)b, and color(green)c`
`color(white)("XXX")x=(-color(blue)33+-sqrt(color(blue)33^2-4 * color(red)(""(-7)) * color(green)(10)))/(2 * color(red)(""(-7)))`

`color(white)("XXX"x)=(-33+-sqrt(1369))/(-14)`

`color(white)("XXX"x)=(-33+-37)/(-14)`

`color(white)("XXX"x)=4/(-14)color(white)("xxxx")or color(white)("xxxx")(-70)/(-14)`

`color(white)("XXX"x)=-2/7color(white)("xxxx")orcolor(white)("xxxx")+5`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

As previously noted these values can be more easily derived from the original factored form, since
if `(x-5)(-7x-2)=0`
then
#{: ("either",x-5=0," or ",-7x-2=0), (,rarrx=+5,,rarrx=-2/7) :}#