Question

Найдите значение аргумента при которых значение функции y= x(в квадрате) - 5 равна 0; 4; - 5.​​

Answer 1

$\displaystyle\bf\\y=x^{2} -5\\\\1)\\\\y=0\\\\0=x^{2} -5\\\\x^{2} =5\\\\x_{1,2} =\pm \ \sqrt{5} \\\\\boxed{\boxed{x_{1} =-\sqrt{5} \ \ ; \ \ x_{2} =\sqrt{5} }}\\\\\\2)\\\\y=4\\\\4=x^{2} -5\\\\x^{2} =9\\\\x_{1,2} =\pm \ \sqrt{9} =\pm \ 3\\\\\boxed{\boxed{x_{1} =-3 \ \ ; \ \ x_{2} =3}}\\\\\\3)\\\\y=-5\\\\-5=x-5\\\\\boxed{\boxed{x=0}}$