Solve without using quadratic formula: $\frac{4}{3x+3} = \frac{12}{x^2 - 1}$. Is there a way to solve this without using the quadratic formula? The quadratic formula is one of my biggest weaknesses, and I would gladly appreciate an easier way to solve this equation.
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3$\begingroup$ the phrase, "the quadratic formula is one of my biggest weaknesses, and I would gladly appreciate an easier way to solve this equation" is really unacceptable at any level of mathematical education. please, make yourself a favour, study and try to understand every aspect of the quadratic formula. solve as many exercises as you can using it, instead of trying to avoid it. You are not going anywhere without it. $\endgroup$– KonKanApr 13, 2016 at 23:45
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3 Answers
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Perhaps something like this: \begin{eqnarray*} \frac{4}{3x+3} & = & \frac{12}{x^{2}-1}\\ \frac{4}{3\left(x+1\right)} & = & \frac{12}{\left(x+1\right)\left(x-1\right)}\\ \frac{4}{3} & = & \frac{12}{\left(x-1\right)}\\ x & = & 10 \end{eqnarray*}
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3$\begingroup$ P.S. and lastly verify that $x=10$ does not make any denominator zero. $\endgroup$– peterwhyApr 14, 2016 at 0:03
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In general, you'll need to use the quadratic formula (or something equivalent like completing the square). If a problem forbids the use of a general technique like this, it's a sign to look for some additional structure.
In this problem, it's that the denominators have a factor in common: $$ 3x+3 = 3(x+1) \quad \textrm{and} \quad x^2-1=(x-1)(x+1) $$
Multiplying both sides of the equation by $(x+1)$ gives $ \frac{4}{3} = \frac{12}{x-1} $, which is solvable through the usual means.
EDIT: in response to "the quadratic formula is one of my biggest weaknesses". Avoiding it only makes it more of a weakness; the best way to become more comfortable with a concept is to use it more often.
answered Apr 13, 2016 at 23:41
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You are supposed to factor the difference of squares in the denominator. Then the problem is easy.
answered Apr 13, 2016 at 23:40
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$\begingroup$ I assumed it was himework and so did not answer the question directly. The OP should be able to find the answer by doing what is suggested. $\endgroup$ Apr 14, 2016 at 0:20