How do you solve #49x^2+3=21#?

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Answer 1 · 2016-10-05T17:11:26

#= (+-3sqrt2)/7#

The first thing to notice is that this is a quadratic equation because it has #x^2#

To solve a quadratic, equation, we would usually make make it equal to 0.

However in this case, there is no #x# term so we can use the method of finding a square root.

#49x^2+3 = 21#

#49x^2 = 18#

#x^2 = 18/49" "larr " isolate " x^2#

#x = +-sqrt(18/49)" "larr# there are 2 square roots to consider

# =+-sqrt((2xx3xx3)/7^2)" "larr# find prime factors of 18

#= (+-3sqrt2)/7#