1)5х (3х+7) + (4х+1)²= -19х+63
2) Приведите уравнение 5х (3х+7) + (4х+1)²= -19х+63 к виду х²+рх+q=0 и укажите значения p и q Пожалуйста, помогите))))
5х (3х+7) + (4х+1)²= -19х+63
15x^2+35x+16x^2+8x+1+19x-63 = 0
31x^2 + 62x - 62 = 0 :| 31
x^2+2x-2 = 0
p = 2
q=-2
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5х (3х+7) + (4х+1)²= -19х+63
15x^2+35x+16x^2+8x+1+19x-63 = 0
31x^2 + 62x - 62 = 0 :| 31
x^2+2x-2 = 0
p = 2
q=-2