# How do you solve 33x^2 - x - 14 = 0?

Apr 7, 2017

For the equation: $33 {x}^{2} - x - 14 = 0$

$x = - \frac{7}{11} \mathmr{and} x = \frac{2}{3}$

#### Explanation:

Given: $33 {x}^{2} - x - 14 = 0$

Since there is an ${x}^{2}$ we know there will be two factors for this equation so we can start setting up the factor brackets:

$\left(x \ldots\right) \left(x \ldots\right) = 0$

We then need to find factors of both $33$ and $14$ that will add or subtract to leave a value of $- 1$ which is the multiplier of the middle or $x$ term.

Factors of $33$ are 33*1; 11*3; and that's it.

Factors of $14$ are 14*1; 7*2; and that's it.

Addition or subtraction of factors $33 \cdot 1 \mathmr{and} 14 \cdot 1$ will not give us $- 1$.

Addition or subtraction of factors $11 \cdot 3 \mathmr{and} 7 \cdot 2$ may give us $- 1$.

Place the factors into the brackets:

$\left(11 x \ldots 7\right) \left(3 x \ldots 2\right) = 0$

We can see that $11 \cdot 2 = 22 \mathmr{and} 7 \cdot 3 = 21$ which is needed to give the $- 1$ we need, if the first is negative and the second is positive.

We also know that one factor will contain a positive integer and the other will contain a negative integer because the $14$ is negative.

Inserting the signs: $\left(11 x + 7\right) \left(3 x - 2\right) = 0$

Now we can solve for the two factors of $x$:

11x+7=0; 11x=-7; x=-7/11

3x-2=0; 3x=2; x=2/3

To check, substitute answers into the $g i v e n$ equation:

$33 {x}^{2} - x - 14 = 0$

$33 {\left(\frac{2}{3}\right)}^{2} - \frac{2}{3} - 14 = 0$

$33 \left(\frac{4}{9}\right) - \frac{2}{3} - 14 = 0$

$\cancel{33} {\left(\frac{4}{\cancel{9}}\right)}^{2} - 14 \frac{2}{3} = 0$

$\frac{44}{3} - 14 \frac{2}{3} = 0$