What is the product of the roots of #ax^2+bx+c = 0" "# where #a != 0# ?

1 Answer
May 15, 2016

Answer:

#c/a#

Explanation:

Method 1 - Direct multiplication

#(-b-sqrt(b^2-4ac))/(2a) * (-b+sqrt(b^2-4ac))/(2a)#

#=((-b)^2-(sqrt(b^2-4ac))^2)/(4a^2)#

#=(b^2-(b^2-4ac))/(4a^2)#

#=(4ac)/(4a^2)#

#=c/a#

#color(white)()#
Method 2 - Think about the roots

Call the two values of this expression #r_1# and #r_2# - the roots of the quadratic equation #ax^2+bx+c = 0#.

Then we have:

#ax^2+bx+c#

#= a(x-r_1)(x-r_2)#

#= a(x^2-(r_1+r_2)x+r_1 r_2)#

#=ax^2-a(r_1 + r_2)x + a r_1 r_2#

Equating coefficients we find:

#c = a r_1 r_2#

So dividing both sides by #a# we find:

#r_1 r_2 = c/a#