Question

625=(35-y)^2+ y^2, помогите с решением

Answer 1

$625 = (35 - y {)}^{2} + {y}^{2} \ 625 = {35 }^{2} - 70y + {y}^{2} + {y}^{2} \ 2 {y}^{2} - 70y + 1225 - 625 = 0 \ 2 {y}^{2} - 70y + 600 = 0 \ {y}^{2} - 35y + 300 = 0$
$D=b^{2}-4ac=1225-4cdot 300=1225-1200=25\y_1 = frac{-b+sqrt{D}}{2a}=frac{35+5}{2}=20\y_2 =frac{-b-sqrt{D}}{2a}=frac{35-5}{2}=15$
Ответ: $y_1 = 20; :: y_2 =15$

Answer 2

625 = (35-y)^2+y^2;
625 = 1225 - 70y + y^2 + y^2;
625 - 1225 + 70y - 2y^2 = 0;
-600 + 70y - 2y^2 = 0;
-2y^2 + 70y - 600 = 0; | : (-2)
y^2 - 35y + 300 = 0
a = 1; b = -35; c = 300;
D = b^2 - 4ac;
D = 1225 - 1200 = √25 = 5
y1,2 = (-b ± √D)/2a
y1 = (35 + 5)/2 = 20
y2 = (35 - 5)/2 = 15
Ответ: 15; 20.