A nice Radical Equation | Math Olympiad | Algebra Problem | Find Real Value of "x" ?

[@manjunathaba9747]

Nice

[@superacademy247]

Thanks

[@RashmiRay-c1y]

Let sqrt(x)=t and 13=a. (t >0) Then t^2-a=sqrt(t+a). Squaring, a^2-a(2t^2+1)+(t^2-t)=0. Solving for a, a = t^2+t+1, t^2-t. The first, setting a = 13, gives t^2+t-12=0 . So t=3 and x =9, which is not allowed. Again, the second is t^2-t-13=0 which gives t = 1/2[sqrt(53)+1] and hence x = 1/2[27+sqrt(53)], which is the answer.

[@rainerzufall42]

27^2 - 4 * 13^2 = 27^2 - 26^2 = 27 + 26 = 53. No need to calculate 729 and 676, because (27 - 26) = 1!

[@woobjun2582]

Thanks alot. Otherwise, let sqrt(x)= t and squaring both sides, (t^2 -13)^2 = 13 + t, and expand, rearrange t^4 -26t^2 -t +156 = 0. By synthetic divisions, (t -3)(t -4)(t^2 -t -13) =0, which results in (1) t = 3 = sqrt(x); x = 9 (< 13, *rejected) (2) t = -4 = sqrt(x); * rejected (3) t^2 -t -17 = 0 t=(1 -sqrt(53))/2 *rejected thus t= (1+sqrt(53))/2 = sqrt(x) x = (27+sqrt(53))/2

[@marceloboda9953]

🇩🇴🇺🇸🇩🇴🇺🇸

[@edelschrott]

I don't know what I did, but my solution was x=x+338 hahaha

[@superacademy247]

That solution doesn't exist unfortunately

[@rainerzufall42]

x = 9.85995, don't be lazy! BZW. sqrt(53) approx.= 7.28011 is a much better approximation than 7.2801... [7.280109889...] That means, x = 17.140055 or x = 9.859945.