How do you solve #4x - 9x ^ { 1/ 2} + 5= 0#?

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Answer 1 · 2016-11-25T12:16:39

#x = 25/16# and #x = 1#

First, factor this equation to give:

#(4x^(1/2) - 5)(x^(1/2) - 1) = 0#

Next, solve for #(4x^(1/2) - 5)#:

#((4x^(1/2) - 5)(x^(1/2) - 1))/(x^(1/2) - 1) = 0/(x^(1/2) - 1)#

#((4x^(1/2) - 5)cancel((x^(1/2) - 1)))/cancel(x^(1/2) - 1) = 0#

#4x^(1/2) - 5 = 0#

#4x^(1/2) - 5 + 5 = 0 + 5#

#4x^(1/2) = 5#

#(4x^(1/2))/4 = 5/4#

#x^(1/2) = 5/4#

#x^(1/2) * x^(1/2) = 5/4 * 5/4#

#x = 25/16#

Finally, solve for #(x^(1/2) - 1)#

#((4x^(1/2) - 5)(x^(1/2) - 1))/(4x^(1/2) - 5) = 0/(4x^(1/2) - 5)#

#(cancel((4x^(1/2) - 5))(x^(1/2) - 1))/cancel(4x^(1/2) - 5) = 0#

#x^(1/2) - 1 = 0#

#x^(1/2) - 1 + 1 = 0 + 1#

#x^(1/2) = 1#

#x^(1/2) * x^(1/2) = 1 * 1#

#x = 1#