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

$= (+-3sqrt2)/7$

How do you solve 49x^2+3=2149x^2+3=21?