Can You Solve This Radical Equation?

[@johnlv12]

amazing

[@RashmiRay-c1y]

The equation can be rewritten as √(3-x)= x^2+2x-3. Let 3=a. Then, squaring, we get a^2-a[2(x^2+2x)+1] + [(x^2+2x)^2+x]=0 > a=3 = 1/2[2(x^2+2x)+1 +/-(2x+1)]. We therefore get two quadratic equations: (I) x^2+3x-2=o > x= 1/2[-3 +/- √17] and (II) x^2+x-3=0 > x=1/2[-1+/-√13]. We can verify that the two valid solutions are x=-1/2[√17 +3], 1/2[√13 -1].

[@9허공]

let t = √(3 - x) , then √(3 - x) = (x - 1)(3 + x) => t = (2 - t^2)(6 - t^2)
=> t^4 - 8t^2 - t + 12 = 0
(1st method)
t^4 - 8t^2 - t + 12 = 0 => (t^2 + at + b)(t^2 - at +c) = 0 => a =1, b =-3, c = -4
=> (t^2 + t - 3)(t^2 - t - 4) = 0
(2nd method)
t^4 - 8t^2 - t + 12 = 0 => (t2 + a/2)^2 - b(t + c/2)^2 = 0 => a =-7, b = 1, c = 1
=> (t^2 - 7/2)^2 - (t + 1/2)^2 = (t^2 - 7/2 + t + 1/2)(t^2 -7/2 - t - 1/2)
=> (t^2 + t - 3)(t^2 - t - 4) = 0

[@Fjfurufjdfjd]

Πρεπει χ=/-3 και χ< =3. Και εχω (χ+3)[(3-χ)^(1/2)]/(χ+3)+(χ+3)=χ(χ+3) ... (3-χ)^(1/2)=χ^2+2χ-3. Θετω 3-χ=y^2 και εχω τελικα (χ-y)(χ+y+1)=0 χ=[-1+-(13)^1/2]/2 χ=[-3+-(17)^1/2]/2

[@RajeshKumar-wu7ox]

X=-2,-4,-3

[@RealQinnMalloryu4]

(9 ➖ x^2)/{6+6 ➖ }{x+x ➖} {x^0+x^0➖}/{6+x^2}=x^1/6x^2 x^1/3^2x^2 x^1/1^1x^2 1x^2 (x ➖ 2x+1) .