How do you solve # x = sqrtx + 6#?

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Answer 1 · 2015-07-20T08:24:46

Substitute #t = sqrt(x)# to get a quadratic in #t#. Solve for #t#, then square to find #x = 9#

Let #t = sqrt(x)#. Note #t >= 0#.

Then #t^2 = t + 6#

Subtract #t+6# from both sides to get:

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

So #t = 3# or #t = -2#.

Discard #t = -2# since #t# is the non-negative square root of #x#.

So #t = 3# and #x = t^2 = 3^2 = 9#