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

2 Answers
May 1, 2017

Answer:

#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)#

May 1, 2017

Answer:

#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#

Tony B