How do you solve #12y ^ { 2} + 15y = 0#?

Question

592 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-13T14:37:35

See a solution process below:

First, factor the left side of the equation as:

#(3 * 4)y^2 + (3 * 5)y = 0#

#3y(4y + 5) = 0#

Now, separately, equate each term to #0# and solve for #y#:

Solution 1)

#3y = 0#

#(3y)/color(red)(3) = 0/color(red)(3)#

#(color(red)(cancel(color(black)(3)))y)/cancel(color(red)(3)) = 0#

#y = 0#

Solution 2)

#4y + 5 = 0#

#4y + 5 - color(red)(5) = 0 - color(red)(5)#

#4y + 0 = -5#

#4y = -5#

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

#(color(red)(cancel(color(black)(4)))y)/cancel(color(red)(4)) = -5/4#

#y = -5/4#

The solutions are: #y = 0# and #y = -5/4#