Before watching: Alright, so we know that X cannot equal 0 (otherwise we have 1/0 on the left). So this means that we can *multiply both sides by X* without losing information. -> 1 + (x^2)/2 = 3x From here we can multiply by 2 to make things a bit easier on ourselves, and subtract 6x from both sides after that: -> 2 + x^2 = 6x.-> 2+x^2-6x = x^2-6x+2=0 (Quick tip: if the coefficient for your x^1 term is negative, it makes it more likely that you'll have real roots). And from HERE, we do our quadratic equation, giving us: (6±√28)/2 = (6±2√7)/2 = 3±√7 Our "roots," and thus the solutions for X in this equation, are 3±√7 (If your test/teacher insist on having you write this without the square root symbol, they'd better either allow calculators or have a table of square roots available. Those roots would then come out to approx. 5.64575 and 0.35425; you may truncate these however you wish)
Before watching:
Alright, so we know that X cannot equal 0 (otherwise we have 1/0 on the left). So this means that we can *multiply both sides by X* without losing information.
-> 1 + (x^2)/2 = 3x
From here we can multiply by 2 to make things a bit easier on ourselves, and subtract 6x from both sides after that:
-> 2 + x^2 = 6x.-> 2+x^2-6x = x^2-6x+2=0
(Quick tip: if the coefficient for your x^1 term is negative, it makes it more likely that you'll have real roots).
And from HERE, we do our quadratic equation, giving us:
(6±√28)/2 = (6±2√7)/2 = 3±√7
Our "roots," and thus the solutions for X in this equation, are 3±√7
(If your test/teacher insist on having you write this without the square root symbol, they'd better either allow calculators or have a table of square roots available. Those roots would then come out to approx. 5.64575 and 0.35425; you may truncate these however you wish)