Question

How do you find the roots, real and imaginary, of `y= 2 (x + 1) (x - 4)` using the quadratic formula?

Answer 1

`x=-1`

`x=4`

Explanation:

You don't need a quadratic formula for this, since it is in the factored form, you just need to set each expression in the brackets equal to zero, and then find the roots.

`x+1=0`
`color(red)(x=-1)`

`x-4=0`
`color(red)(x=4)`

Answer 2

`x=4 and -1`

Explanation:

As we are instructed to use the formula it is perhaps better to change the given equation into the format of `y=ax^2+bx+c`

Not only that, I suspect that part of the agenda of this question is to see if you can 'handle' expansion of brackets.

Multiplying everything inside the right brackets by everything in the left.

`y=2color(red)((x+1))color(green)((x-4))`

`y=2color(green)([color(white)(2/2)color(red)(x)(x-4)color(red)(+1)(x-4)" "])`

`y=2(x^2-4x+x-4)" "->" "y=2x^2-6x-8`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Thus we have;`" "a=2"; "b=-6"; "c=-8`

So `x=(-b+-sqrt(b^2-4ac))/(2a)` becomes:

`x=(+6+-sqrt((-6)^2-4(2)(-8)))/(2(2))`

`x=(+6+-sqrt(36+64))/4`

As the content of the root (determinate) is not negative there are no imaginary roots. That is, in this case, the graph crosses the axis.

Note: if the determinate is 0 then the x-axis is tangential to the max/min

`x=3/2+-5/2`

`x=-2/2->-1`
`x= 8/2->4`