Каковы корни уравнения 4x ^ {2} = 2+ 7x?

Ответ:

# Х = 2 #

# Х = -1/4 #

Объяснение:

Дано -

# 4x ^ 2 = 2 + 7x #

# 4x ^ 2-7x-2 = 0 #

# Х ^ 2-7 / 4x-2/4 = 0 #

# Х ^ 2-7 / 4x-1/2 = 0 #

# Х ^ 2-7 / 4x = 1/2 #

# x ^ 2-7 / 4x + 49/64 = 1/2 + 49/64 = (32 + 49) / 64 = 81/64 #

# (Х ^ 2-7 / 8) ^ 2 = 81/64 #

# (Х-7/8) = + - SQRT (81/64) #

# (Х-7/8) = + - 9/8 #

# Х = 9/8 + 7/8 = (9 + 7) / 8 = 16/8 = 2 #

# Х = 2 #

# Х = -9/8 + 7/8 = (- 9 + 7) / 8 = -2 / 8 = -1 / 4 #

# Х = -1/4 #