# If 3 times the square of an integer is added to 1 times the integer, the result is 2. Find the...

## Question:

If {eq}3 {/eq} times the square of an integer is added to {eq}1 {/eq} times the integer, the result is {eq}2 {/eq}.

Find the integer.

## Algebraic Equations: One Unknown

Algebraic equations involving just one unknown are typically the simplest equations we deal with. The idea is to rewrite the relationships described in the problem in terms of mathematical equations and then perform necessary operations to rearrange them and solve for the unknown, say, x. In this specific problem the tool required to solve this equation is the quadratic formula.

For an equation in a form {eq}ax^2 + bx + c = 0 {/eq}, the two solutions would be given by the following formula:

{eq}x_{1, 2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} {/eq}.

The goal of this problem is to identify an unknown integer, x.

Within this problem, we're given that 3 times of the square of an integer, that is, {eq}3x^2 {/eq}, is added to the integer, x, the result is 2. Writing this as an equation, we obtain:

{eq}3x^2 + x = 2...(1) {/eq}.

Let's rearrange it into a quadratic form:

{eq}3x^2 + x - 2 = 0...(2) {/eq}

Apply the quadratic formula knowing that here we have: a = 3, b = 1, c = -2:

{eq}x_{1, 2}=\frac{-1 \pm \sqrt{1^2 - 4\cdot 3\cdot (-2)}}{2\cdot 3} = \frac{-1 \pm 5}{6} {/eq}

The first solution would be:

{eq}x_1 = \frac{-1 + 5}{6} = \frac{2}{3} {/eq}, and the second solution:

{eq}x_2 = \frac{-1 - 5}{6} = -1 {/eq}.

Only the second solution is an integer. Therefore, x = -1.