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

1 Answer
Feb 11, 2018

Answer:

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#

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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) :}#