Melyek a 4x ^ {2} = 2+ 7x egyenlet gyökerei?

$Melyek a 4x ^ {2} = 2+ 7x egyenlet gyökerei?$

Válasz:

# X = 2 #

# X = -1/4 #

Magyarázat:

Adott -

# 4x ^ 2 = 2 + 7x #

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

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

# X ^ 2-7 / 4x-1/2 = 0 #

# X ^ 2-7 / 4x = 1/2 #

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

# (X ^ 2-7 / 8) ^ 2 = 81/64 #

# (X-7/8) = + - sqrt (81/64) #

# (X-7/8) = + - 9/8 #

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

# X = 2 #

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

# X = -1/4 #