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 Question from Nick: I'm an 8th grader and am quite confused with this equation: 3x^2-19x-14=0 The answer is: -2/3 , 7 But I'm not sure how to solve for it.

Hi Nick,

You have equation with a quadratic expression with integer coefficients on one side and zero on the other side. Your instructions are to solve for the variable. Whenever you have this situation you should first try to factor the quadratic. So the question is, can you find integers $a, b, c$ and $d$ so that

$3x^2 - 19x - 14 = (ax + b)(cx + d).$

If you knew $a, b, c$ and $d$ you could expand the right side to verify that your factoring was correct. I would start expanding the right side with

$ax \times cx = a \times b x^2$

and this must be $3x^2$ so $a \times c = 3.$ Since $a$ and $c$ are integers one of them must be 3 and the other 1. Thus we have

$3x^2 - 19x - 14 = (3x + b)(x + d).$

I would continue expanding the right side with the constant term $b \times d.$ This must be the constant term on the left side so $b \times d = -14.$ This tells me two things, first since $-14$ is negative $b$ and $d$ have opposite signs. Also since $14$ can only be factored as $14 \times 1$ or $7 \times 2.$ Hence you can write all the possibilities for the factorization

$3x^2 - 19x - 14 = (3x + b)(x + d).$

Writing $14$ as $14 \times 1$ gives

\begin{eqnarray*}
(3x + 1)&(&x - 14) \\(3x - 1)&(&x + 14)\\(3x + 14)&(&x - 1)\\(3x - 14)&(&x + 1)
\end{eqnarray*}

and writing $14$ as $7 \times 2$ gives

\begin{eqnarray*}
(3x + 2)&(&x - 7) \\(3x - 2)&(&x + 7)\\(3x + 7)&(&x - 2)\\(3x - 7)&(&x + 2).
\end{eqnarray*}

You can now expand the eight possibilities and see if there is one that gives $3x^2 - 19x - 14.$

See if you can complete the problem now and write back if you need more assistance.

Penny

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