How do you solve #5(sqrt x) - x =6#?

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Answer 1 · 2015-07-20T23:31:25

Let #t = sqrt(x)#, solve the resulting quadratic in #t#,

then derive #x=4# or #x=9#.

Let #t=sqrt(x)#.

Note #t >= 0# since #sqrt# denotes the non-negative square root.

Then #5t - t^2 = 6#

Add #t^2-5t# to both sides to get:

#0 = t^2-5t+6 = (t-2)(t-3)#

So #t=2# or #t=3#. These both satisfy #t >= 0# so are valid solutions.

So #x = 2^2 = 4# or #x = 3^2 = 9#