Asked by alice
Find all values of $a$ that satisfy the equation
\[\frac{a}{3} + 1 = \frac{a + 3}{a} - \frac{a^2 + 2}{a}.\]
\[\frac{a}{3} + 1 = \frac{a + 3}{a} - \frac{a^2 + 2}{a}.\]
Answers
Answered by
oobleck
what is so hard about just using regular math notation?
What we were given was
\[\frac{a}{3} + 1 = \frac{a + 3}{a} - \frac{a^2 + 2}{a}.\]
a/3 + 1 = (a+3)/a - (a^2+2)/a
multiplying through by 3a, that gives
a^2 + 3a = 3(a+3) - 3(a^2+2)
a^2 + 3a = 3a+9-3a^2-6
4a^2 = 3
a = ±√3/2
What we were given was
\[\frac{a}{3} + 1 = \frac{a + 3}{a} - \frac{a^2 + 2}{a}.\]
a/3 + 1 = (a+3)/a - (a^2+2)/a
multiplying through by 3a, that gives
a^2 + 3a = 3(a+3) - 3(a^2+2)
a^2 + 3a = 3a+9-3a^2-6
4a^2 = 3
a = ±√3/2
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