Question

Решите методом через модуль, пожалуйста

Answer 1

$\displaystyle\bf\\x^{4} =(4x-5)^{2} \\\\x^{2} =|4x-5|\\\\\\\left[\begin{array}{ccc}x^{2} =4x-5\\x^{2} =-4x+5\end{array}\right \ \ \ \Rightarrow \ \ \left[\begin{array}{ccc}x^{2} -4x+5=0\\x^{2} +4x-5=0\end{array}\right \\\\\\1) \ x^{2} -4x+5=0\\\\D=(-4)^{2} -4\cdot 5=16-20=-4 < 0\\\\x\in \oslash\\\\2) \ x^{2} +4x-5=0\\\\Teorema \ Vieta:\\\\x_{1} + x_{2} =-4\\\\x_{1} \cdot x_{2} =-5\\\\\boxed{x_{1} =1 }\ \ \ ; \ \ \ \boxed{x_{2} =-5}$