The Best Substitution to Solve Radical Equations Faster!

[@RyanLewis-Johnson-wq6xs]

It’s in my head.

[@RyanLewis-Johnson-wq6xs]

x=11+Sqrt[(5x-56)/(x-13)(x-14)] x=13±Sqrt[3] x=11.5±0.5Sqrt[3]i=(23±Sqrt[3]i)/2

[@潘博宇-k4l]

X>14, X=13+(3)^(1/2).

[@RashmiRay-c1y]

Let x-11=a. Therefore, a = [(5a-1)/{(a-2)(a-3)}]^1/2 and squaring, a^4-5a^3+6a^2-5a+1=0 which gives a^2+1/a^2+2 -5(a+1/a) +4=0 which in turn gives a+1/a=1,4. a+1/a=1 does not give real x. If a+1/a=4, a=2+/-√3 and x = 13 +/- √3.

[@Quest3669]

X= 13+-√3 real valid sons.

[@Usershima-c4v]

X=(23+√5)/2 X=13+√3

[@Fjfurufjdfjd]

Στο R χ=13+(3)^(1/2) ή χ=13-(3)^(1/2).

[@adribber]

x = 11 + √{(5x - 56)/(x - 13)(x - 14)} (x ∈ R) x - 11 = √{(5x - 56)/(x - 13)(x - 14)} put, y = x - 11 y = √{(5y - 1)/(y - 2)(y - 3)} y² = (5y - 1)/(y - 2)(y - 3) y²(y - 2)(y - 3) = 5y - 1 y²(y - 2)(y - 3) - 5y + 1 = 0 y²(y² - 5y + 6) - 5y + 1 = 0 y⁴ - 5y³ + 6y² - 5y + 1 = 0 y² - 5y + 6 - 5/y + 1/y² = 0 (y² +2 + 1/y²) - 5(y + 1/y) + 4 = 0 (y + 1/y)² - 5(y + 1/y) + 4 = 0 put, t = y + 1/y t² - 5t + 4 = 0 (t - 1)(t - 4) = 0 ∴ t = 1 or 4 Case 1 : t = 1 y + 1/y = 1 y² - y + 1 = 0 D = (-1)² - 4*1*1 = - 3 < 0 rejected Case 2 : t = 4 y + 1/y = 4 y² - 4y + 1 = 0 y = [4 ± √{(-4)² -4*1*1}] / 2*1 = 2 ± √3 x = y + 11 = 11 + 2 ± √3 = 13 ± √3 ∴ x = 13 + √3 , 13 - √3