How do you solve #64x^2 - 144x + 81 = 0#?

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Answer 1 · 2015-05-18T00:18:08

We can use Bhaskara here.

#(-b+-sqrt(b^2-4ac))/(2a)#

where #a=64#, #b=-144# and #c=81#.

#(144+-sqrt(144^2-4(64)(81)))/128#

#(144+-sqrt(20736-20736))/128#

#(144+-sqrt(0))/128#

#144/128=color(green)(9/8)#

Your quadratic function is tangent to axis #x#, touching it only in the point with #x# coordinate #(9/8)#. So, we can say that #x_1=x_2=9/8#

Answer 2 · 2015-05-18T00:20:30

#y = (8x - 9)^2 = 64x^2 - 144x + 81.# = 0, then:

#(8x - 9) = 0 -> x = 9/8.# There is double root.