# The product of two integers is -120 and their sum is 2. How can you find the larger of the two numbers?

Feb 8, 2017

Answer $D$ would be correct.

#### Explanation:

Let the two integers be $x$ and $y$.

Then

$\left\{\begin{matrix}x y = - 120 \\ x + y = 2\end{matrix}\right.$

I would recommend solving through substitution.

$y = - \frac{120}{x} \to x - \frac{120}{x} = 2$

${x}^{2} - 120 = 2 x$

${x}^{2} - 2 x - 120 = 0$

You could use factoring, completing the square or quadratic formula to solve. I'll use completing the square.

${x}^{2} - 2 x = 120$

$1 \left({x}^{2} - 2 x + 1 - 1\right) = 120$

$1 \left({x}^{2} - 2 x + 1\right) - 1 = 120$

${x}^{2} - 2 x + 1 = 121$

${\left(x - 1\right)}^{2} = 121$

$x - 1 = \pm 11$

$x = - 11 + 1 \mathmr{and} 11 + 1$

$x = - 10 \mathmr{and} 12$

Resubstitute into the initial equation:

$x + y = 2 \to - 10 + y = 2 \mathmr{and} 12 + y = 2 \to y = 12 \mathmr{and} y = - 10$

Therefore, the larger number is $+ 12$.

Hopefully this helps!

Feb 8, 2017

D) +12

#### Explanation:

There are many combinations of numbers which multiply to give $- 120$

We know one must be positive and one must be negative.

For example $1 \times - 120 = - 120 , \text{ } \mathmr{and} - 2 \times 60 = - 120$

However these values have quite a big sum. (If we think about the absolute value and do not look at the signs)

$= - 120 + 1 = - 119 \text{ } \mathmr{and} 60 + \left(- 2\right) = 58$

To have a sum of only $+ 2$ means that the two numbers are of a very similar size, with the positive one being slightly bigger than the negative one.

Look for 'middle' factors - ie, those very close to the square root.

$\sqrt{120} = 10.95$

Find factors of $120$ on either side of $10.95 \left(\approx 11\right)$

10 xx 12 = 120!

The $12$ must be positive and the $10$ negative.

$- 10 \times 12 = - 120 \text{ " and" } - 10 + 12 = + 2$

The larger one is $+ 12$