Question

Будь ласка розв'яжіть рівняння.
${3x}^{4} - {2x}^{2} - 40 = 0$
та
$( {x}^{2} - x)^{2} - 11( {x}^{2} - x) + 18 = 0$
​

Answer 1

$\displaystyle\bf\\1)\\\\3x^{4} -2x^{2} -40=0\\\\x^{2} =m \ , \ m > 0\\\\3m^{2} -2m-40=0\\\\D=(-2)^{2} -4\cdot 3\cdot(-40)=4+480=484=22^{2} \\\\\\m_{1}=\frac{2+22}{6} =4\\\\\\m_{2} =\frac{2-22}{6} =-3\frac{1}{3} < 0-neyd\\\\x^{2} =4\\\\x_{1} =-2 \ \ ; \ \ x_{2} =2\\\\Otvet \ : \ - 2 \ ; \ 2$

$\displaystyle\bf\\2)\\\\\Big(\underbrace{x^{2} -x}_{m}\Big)^{2} -11\Big(\underbrace{x^{2} -x}_{m}\Big)+18=0\\\\x^{2} -x=m\\\\m^{2} -11m+18=0\\\\D=(-11)^{2} -4\cdot 18=121-72=49=7^{2}\\\\\\m_{1} =\frac{11-7}{2} =2\\\\\\m_{2} =\frac{11+7}{2}=9\\\\1)\\\\m=2\\\\x^{2} -x=2\\\\x^{2} -x-2=0\\\\Teorema \ Vieta:\\\\x_{1} + x_{2}=1\\\\x_{1} \cdot x_{2} =-2\\\\x_{1} =2 \ \ \ ; \ \ \ x_{2} =-1$

$\displaystyle\bf\\2)\\\\x^{2} -x=9\\\\x^{2} -x-9=0\\\\D=(-1)^{2} -4\cdot (-9)=1+36=37\\\\\\x_{3} =\frac{1-\sqrt{37} }{2} \\\\\\x_{4} =\frac{1+\sqrt{37} }{2} \\\\\\Otvet \ : \ 2 \ \ ; \ \ -1 \ \ ; \ \ \frac{1-\sqrt{37} }{2} \ \ ; \ \ \frac{1+\sqrt{37} }{2}$