Find three consecutive integers such that the square of the first, increased by the last, is 32.

In order to find the 3 consecutive digits here, we need to note that consecutive numbers are numbers that appear as the number right above each other.

For example: 2, 3, 4 are consecutive, as are -10, -9, -8.

We can assign the first term a variable, let's do x. Since we know the next two terms are consecutive, we can define them with [tex]x+1[/tex] and [tex]x+2[/tex].

[tex]x, (x+1), (x+2)[/tex]

We also know that the first number squared, increased by the last, is 32. This can be modeled by the equation

[tex]x^2 + (x+2) = 32[/tex]

Let's solve for x in that equation by using the XBOX Method in quadratics.

[tex]x^2 + x + 2 = 32[/tex]
[tex]x^2 + x - 30 = 0[/tex]
The product of the two roots will be c (-30) and their sum will be b (1).
[tex]-5 \cdot 6 = -30[/tex]
[tex]-5 + 6 = 1[/tex]
[tex](x-5)(x+6)[/tex]
Zeroes of the function: [tex]5, -6[/tex]

Now that we know two values of x that might work, we need to plug them into our equation to test if they actually do work.

5

[tex]5^2 + (5+2) = 32\\\\ 25 + 7 = 32 \ \checkmark[/tex]

-6

[tex]-6^2 + (-6+2) = 32 \\\\ -36 + -4 = 32 \ \times[/tex]

We can see here that -6 won't work as it doesn't satisfy our equation. However, 5 does work. That means our first number is 5, making our next two numbers 6 and 7.

Hence - 5, 6, 7.

Hope this helped!