How do you solve #3x^2+5x=0# using the quadratic formula?

2 Answers
Aug 26, 2015

Answer:

#x=0#
or
#x=-5/3#

Explanation:

Standard quadratic form:
#color(white)("XXXX")ax^2+bx+c=0#

Re-writing the given equation in explicit standard form:
#color(white)("XXXX")3x^2+5x+0=0#
with #a=3#, #b=5#, and #c=0#

The quadratic formula is
#color(white)("XXXX")x = (-b+-sqrt(b^2-4ac))/(2a)#

which, in this case becomes
#color(white)("XXXX")x = (-5+-sqrt(5^2-4(3)(0)))/(2(3)#

#rArr x= 0 or x=-10/6 = -5/3#

Aug 26, 2015

Answer:

The solutions are

#color(blue)(x=0#

#color(blue)(x=-5/3#

Explanation:

The equation #3x^2+5x=0#:
is of the form #color(blue)(ax^2+bx+c=0# where:

#a=3, b=5, c=0#
(the equation lacks a constant term)

The Discriminant is given by:
#Delta=b^2-4*a*c#
# = (5)^2-(4*3*0)#
# = 25 - 0=25#

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

#x = ((-5)+-sqrt(25))/(2*3) = ((-5+-5))/6#

#x=((-5+5))/6, color(blue)(x=0#

#x=((-5-5))/6, x=-10/6 color(blue)(x=-5/3#