How do you solve #15a^2=60a#?

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Answer 1 · 2016-12-18T21:46:05

#a= 0" or "a = 4#

This is a quadratic equation, set it equal to 0.

#15a^2 -60a = 0" "larr# divide by 15 on both sides.

#a^2 -4a = 0" "#factor out the HCF which is #a#

# a(a-4) = 0#

Set each factor equal to 0

#a = 0" or "a-4=0#

#a= 0" or "a = 4#

Answer 2 · 2016-12-18T22:27:00

#color(red)("This a trap: You think that there is only 1 value for "a)#

Very wrong! #" "a=0" or "a=4#

#color(blue)("The trap")#

divide both sides by #a#

#15a=60#

Divide both sides by 15

#a=60/15 = 12/3 = 4#

#a=4#
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#color(blue)("The better way")#

Divide both sides by 15

#a^2=4a#

#=>a^2-4a=0#

#a(a-4)=0#

#a=0" or "a=4#