Question

Найдите меньшее из двух чисел,сумма которых равна 22,а сумма их квадратов - 250

Answer 1

$\left \{ {{x + y = 22} \atop {x^2 + y^2 = 250}} \right. \\\\ \left \{ {{y = 22 - x} \atop {x^2 + (22 - x)^2 = 250}} \right. \\\\ \left \{ {{y = 22 - x} \atop {x^2 + 484 - 44x + x^2 = 250 \ | \ : \ 2}} \right. \\\\ \left \{ {{y = 22 - x} \atop {x^2 - 22x + 242 = 125}} \right. \\\\ \left \{ {{y = 22 - x} \atop {x^2 - 22x + 117= 0}} \right.$

$x^2 - 22x + 117 = 0\\\\ D = 484 - 468 = 16\\\\ x_1 = \frac{22 + \sqrt{16}}{2} = 13, \ y_1 = 22 - 13 = 9\\\\ x_2 = \frac{22 - \sqrt{16}}{2} = 9, \ y_2 = 22 - 9 = 13\\\\ \boxed{\mathbb{OTBET}: 9}$

Answer 2

Составим систему
х+у=22 х=22-у x=9;y=13
=> =>
х^2+у^2=250 2у^2-44у+234=0 y=13;y=9

D=1936-4*2*234=64
y1=(44+8)/4=13 y2=(44-8)/4=9

Ответ:9