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

1 Answer
Mar 25, 2017

Answer:

#x_1=5+0i#

#x_2=2+0i#

Explanation:

#y=(x-5)(x-2)=x^2-7x+10#

Let #y=0#

#x^2-7x+10=0#

Using Descartes' rule for #f(x)#, we know that there are two positive roots. Using it for #f(-x)#, we know that there can be zero or two negative roots.

#Delta=(-7)^2-4(10)(1)=49-40=9#

#x=(7+-sqrt(9))/2=(7+-3)/2=5, 2#

There are no negative roots and a quadratic can only have, at most, two real roots.

So the real values of #x# are #5# and #2#. These are also the complex roots are we can write them as #5+0i# and #2+0i#.