Question

Решите пожалуйста уравнение: 49 y^2+14y+1=25

Answer 1

49у²+14у+1=25
49у²+14у+1-25=0
49у²+14у-24=0
Д=196+4*49*24=4900=70²
у1=
$\frac{ - 14 + 70}{2 \times 49?} = \frac{56}{98 } = \frac{4}{7}$
у2=
$\frac{ - 14 - 70}{2 \times 49} = \frac{ - 84}{98} = - \frac{6}{7}$

Answer 2

$\displaystyle\tt 49y^2+14y+1=25\\\\(7y+1)^2=5^2\\\\(7y+1)^2-5^2=0\\\\ (7y+1-5)(7y+1+5)=0\\\\(7y-4)(7y+6)=0\\\\ 7y-4=0 \ \ \ \ \ \ \ \ 7y+6=0\\7y=4 \ \ \ \ \ \ \ \ \ \ \ \ \ \ 7y=-6\\y=\frac{4}{7} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y=-\frac{6}{7}$

Ответ: $\tt y_1=-\cfrac{6}{7} ; \ \ \ y_2=\cfrac{4}{7}$