Question

x+√x -20=0

x^5-9x^3+20x=0

Answer 1

1)$x + sqrt{x} - 20 = 0\ t^{2} = x\ t^{2} + t - 20 = 0\ D = 1+ 80 = 81 = sqrt{9}\ t_{1} = -5\ t_{2} = 4\ x_{1} = 25\ x_{2} = 16$

2)$x^{5} - 9x^{3} +20x =0\ x(x^{4} - 9x^{2} +20) = 0\ x = 0\ x^{4} - 9x^{2} +20 = 0\ t = x^{2}\ t^{2} - 9t +20 = 0\ D = 81- 80 = 1\ t_{1} = 4\ t_{2} = 5\ x_{1} = 2\ x_{2} = sqrt{5}$

Answer 2

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