How do you solve #3x ^ { 2} - 90x - 192= 0#?

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Answer 1 · 2017-07-15T16:31:33

# x=+32 and -2#

Introduction to the test for divisibility by 3:

The sum of the digits in 90 is #9+0=9# which is divisible by 3 so 90 is divisible by 3.

Using the same method as above: #1+9+2=12# so 192 is divisible by 3.

Factor out the 3

#3(x^2-30x-64)=0#

Divide both sides by 3

#x^2-30x-64=0#

Notice that #2xx32=64" and that "32-2=30#

#(x-32)(x+2)=0#

#=> x=+32 and -2#
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Check:

#x^2-30x-64=0" "->(-2)^2-30(-2)-64" gives "0#

#x^2-30x-64=0" "->(32)^2-30(32)-64" gives "0#