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Normally, the question will be given as Solve 2x^2 - x - 3=0 if the LHS is factorisable. In this case, the question will be Solve, giving your answers to 2 decimal places, 4x^2 - 5x - 12 =0, hinting that the LHS is not factorisable. OR Solve, giving exact answers, (meaning surds), 4x^2 - 5x - 12 = 0.
ah yes olympiad math. The best way to make math a trillion times more complicated then it needs to be. 10 seconds in my head I knew 27 ^ 1/2 is 3(3 ^ 1/2) didn't need to square or cube y just needed to know how to simplify.
x^2-y^2=27,one of them is odd and other is even. xy=18=18×1=9×2=6×3 x^2-y^2=18^2-1^2 or 9^2-2^2 or 6^2-3^2 Here 6^2-3^2 is correct. (x,y)=(6,3) or (-6,-3) x+y=9 or -9.
Naresh, x<1 is not a necessary condition, as x^(1/4) can be negative, but the starting equation must be verified. Only the positive Sqrt(x) can be used as the x^(1/4) must be a real number. For the first root x1<1, the equation works for only the positive x^(1/4). For the second root x2>1, the equation works for only the negative x^(1/4). So both roots x1 and x2 works conditionally.
If you raise both sides to the power of x, then you get 3 = 4^x. Now it is just a simple exponential problem, where you can take the ln of both sides to get ln(3) = x*ln(4) [by logarithm rule that says ln(a^x) = x*ln(a)]. Divide both sides by ln(4) to get x = ln(3) / ln(4).
I am sorry but one of the two solutions is false for x^1/4+x^1/2=1 because x must be less than 1. So the correct solution is only x = (7-3(5^1/2))/2 which is less than 1. The other solution is greater than 1 (not valid).
(11⁻¹)¹/¹¹ = (11)⁻¹/¹¹ . . . but exponent should be +, so add and subtract one . . . 1-(1/11) makes exponent + . . . subtracting 1 restores the balance . . . The exponent becomes 10/11 -1 . . . with base becomes (11) ¹⁰/¹¹ (11)⁻¹ . . . which is (11) ¹⁰/¹¹ all divided by 11
