How do you solve #9u ^ { 2} - 24u + 16=0#?

1 Answer
Jan 23, 2018

#u = 4/3" "# with multiplicity #2#

Explanation:

Note that:

#(A-B)^2 = A^2-2AB+B^2#

In our example, both #9u^2 = (3u)^2# and #16 = 4^2# are perfect squares and we find that putting #A=3u# and #B=4#, the middle term matches #-2AB# ...

#0 = 9u^2-24u+16#

#color(white)(0) = (3u)^2-2(3u)(4)+4^2#

#color(white)(0) = (3u-4)^2#

Hence:

#u = 4/3" "# with multiplicity #2#