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

2 Answers
Aug 13, 2016

Answer:

#x={0,-5/2}#

Explanation:

#ax^2+bx+c=0#

#2x^2+5x=0#

#a=2" ; "b=5" ; "c=0#

#Delta=sqrt(b^2-4*a*c)#

#Delta=sqrt(5^2-4*2*0)#

#Delta=sqrt 5^2#

#Delta=±5#

#x_("1,2")=(-b±Delta)/(2*a)#

#x_1=(-5-5)/(2*2)=-10/4=-5/2#

#x_2=(-5+5)/(2*2)=0#

#x={0,-5/2}#

#"Or simply"#

#2x^2+5x=0#

#2cancel(x^2)=-5cancel(x)#

#2x=-5#

#x=-5/2#

Aug 13, 2016

Answer:

#x=-5/2, 0#

Explanation:

#2x^2+5x=0# is a quadratic equation in the form #ax^2+bx+c#, where #a=2#, #5=b#, and #c=0#.

Quadratic formula

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

Plug the given values into the formula.

#x=(-5+-sqrt((-5)^2-4*2*0))/(2*2)#

Simplify.

#x=(-5+-sqrt(25-0))/4#

Simplify.

#x=(-5+-sqrt 25)/4#

Simplify.

#x=(-5+-5)/4#

Solve for #x_1# and #x_2#.

#color(blue)(x_1=(-5+5)/4)#

#color(blue)(x_1=0/4)#

#color(blue)(x_1=0)#

#color(red)(x_2=(-5-5)/4)#

#color(red)(x_2=-10/4)#

#color(red)(x_2=-5/2)#

#color(purple)(x=-5/2, 0)#