How do you solve #7x^2 = 21x# using the quadratic formula?

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Answer 1 · 2016-05-17T06:37:15

The solutions are:
#x = 3#

#x= 0#

#7x^2 = 21x#

#7x^2 - 21x = 0 #

The equation is of the form #color(blue)(ax^2+bx+c=0# where:

#a=7, b=-21, c=0#
(the expression has no constant term)

The Discriminant is given by:

#Delta=b^2-4*a*c#

# = (-21)^2-(4* 7 * 0)#

# = 441#

The solutions are found using the formula
#x=(-b+-sqrtDelta)/(2*a)#

#x = (-(-21)+-sqrt(441))/(2*7) = (21+-21)/14#

#x=(21+21) / 14 = 42/14 = 3#

#x=(21-21) / 14 = 0/14 = 0#