Rozwiązane

4x(5x²+2x)+3(x²-x)=0 ?



Odpowiedź :

Magda

Odpowiedź:

[tex]4x(5x^2+2x)+3(x^2-x)=0\\\\20x^3+8x^2+3x^2-3x=0\\\\20x^3+11x^2-3x=0\\\\x(20x^2+11x-3)=0\\\\x(20x^2+15x-4x-3)=0\\\\x(5x(4x+3)-(4x+3))=0\\\\x(4x+3)(5x-1)=0\\\\x=0\ \ \ \ \vee\ \ \ \ 4x+3=0\ \ \ \ \ \ \vee\ \ \ \ 5x-1=0\\\\x=0\ \ \ \ \vee\ \ \ \ 4x=-3\ \ /:4\ \ \vee\ \ \ \ 5x=1\ \ /:5\\\\x=0\ \ \ \ \vee\ \ \ \ x=-\frac{3}{4}\ \ \ \ \ \ \ \ \ \ \ \vee\ \ \ \ x=\frac{1}{5}[/tex]

Basetla

[tex]4x(5x^{2}+2x)+3(x^{2}-x) = 0\\\\20x^{3}+8x^{2}+3x^{2}-3x = 0\\\\20x^{3}+11x^{2}-3x = 0\\\\x(20x^{2}+11x-3) = 0\\\\x_{o} = 0\\\\Lub:\\\\20x^{2}+11x-3 = 0\\\\\Delta = b^{2}-4ac = 11^{2}-4\cdot20\cdot(-3) = 121 + 240 = 361\\\\\sqrt{\Delta} = \sqrt{361} = 19\\\\x_1 = \frac{-b-\sqrt{\Delta}}{2a} = \frac{-11-19}{2\cdot20} = \frac{-30}{40} = -\frac{3}{4}\\\\x_2 = \frac{-b+\sqrt{\Delta}}{2a} = \frac{-11+19}{40} = \frac{8}{40} = \frac{1}{5}\\\\x \in \{-\frac{3}{4}, 0, \frac{1}{5}\}[/tex]