How do you solve # x = sqrtx + 6#? Algebra Radicals and Geometry Connections Radical Equations 1 Answer George C. Jul 20, 2015 Substitute #t = sqrt(x)# to get a quadratic in #t#. Solve for #t#, then square to find #x = 9# Explanation: 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# Answer link Related questions How do you solve radical equations? What are Radical Equations? How do you solve radical equations with cube roots? How do you find extraneous solutions when solving radical equations? How do you solve #\sqrt{x+15}=\sqrt{3x-3}#? How do you solve and find the extraneous solutions for #2\sqrt{4-3x}+3=0#? How do you solve #\sqrt{x^2-5x}-6=0#? How do you solve #\sqrt{x}=x-6#? How do you solve for x in #""^3sqrt(-2-5x)+3=0#? How do you solve for x in #sqrt(42-x)+x =13#? See all questions in Radical Equations Impact of this question 1539 views around the world You can reuse this answer Creative Commons License