If 3 times the square of an integer is added to 1 times the integer, the result is 2. Find the integer.
3x^2 + x = 2
3x^2 + x - 1 = 0
no integer solution, (use the formula, it will not come out to be a whole number)
Wait Mr. Reiny, I think there is a typo.
3x^2+x-2=0
So there should be two roots
x= -1
x = 2/3
And since the question is asking for "integer", I guess only -1 works since 2/3 is a fraction which is not an integer.
To solve this problem, let's break it down step by step:
1. Let's assume the integer is represented by the variable 'x'.
2. According to the problem, we need to find the value of 'x' such that when we multiply its square by 3 (3 * x^2) and then add it to the product of the integer and 1 (x * 1), the result is 2. Mathematically, this can be expressed as:
3 * x^2 + x * 1 = 2
3. Now, let's simplify the equation and get rid of any unnecessary terms.
3x^2 + x = 2
4. Rearrange the equation so that it is in standard quadratic form:
3x^2 + x - 2 = 0
5. At this point, we can attempt to factor the quadratic equation or use the quadratic formula to find the values of 'x'.
Since the equation does not easily factor, let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
For our equation, a = 3, b = 1, and c = -2.
6. Plugging these values into the quadratic formula, we have:
x = (-1 ± √(1^2 - 4 * 3 * -2)) / (2 * 3)
7. Simplifying further:
x = (-1 ± √(1 + 24)) / 6
= (-1 ± √25) / 6
= (-1 ± 5) / 6
8. Finally, we have two possible solutions for 'x':
x = (-1 + 5) / 6
= 4 / 6
= 2/3
or
x = (-1 - 5) / 6
= -6 / 6
= -1
Therefore, the two possible integers are 2/3 and -1.