How do you evaluate #20a ^ { 2} - 11a = 3#?

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Answer 1 · 2016-12-11T23:30:49

#a = 3/4# and #a = -1/5#

First, get this quadratic in terms of solving for #0#:

#20a^2 - 11a - 3 = 3 - 3#

#20a^2 - 11a - 3 = 0#

Now we can factor by playing with multipliers for 20 (1x20, 2x10, 4x5, 5x4, 10x2, 20x1) and multipliers for 3 (1x3 and 3x1):

#(4a - 3)(5a + 1) = 0#

Then we can solve each term for #0#:

#4a - 3 = 0#

#4a - 3 + 3 = 0 + 3#

#4a = 3#

#(4a)/4 = 3/4#

#a = 3/4#

and

#5a + 1 = 0#

#5a + 1 - 1 = 0 - 1#

#5a = -1#

#(5a)/5 = -1/5#

#a = -1/5#