Question Find the integers whose sum of squares is 685.

Studdy Solution

STEP 1

Assumptions
1. We are looking for two consecutive integers.
2. The sum of the squares of these integers is 685.
3. We denote the smaller integer as $n$ and the larger integer as $n+1$ , where $n$ is an integer.

STEP 2

Express the given condition using the assumption that the smaller integer is $n$ .
$n^2 + (n+1)^2 = 685$

STEP 3

Expand the square of the second term $(n+1)^2$ .
$n^2 + n^2 + 2n + 1 = 685$

STEP 4

Combine like terms to simplify the equation.
$2n^2 + 2n + 1 = 685$

STEP 5

Subtract 685 from both sides to set the equation to zero.
$2n^2 + 2n + 1 - 685 = 0$

STEP 6

Simplify the equation by combining constants.
$2n^2 + 2n - 684 = 0$

STEP 7

Divide the entire equation by 2 to simplify further.
$n^2 + n - 342 = 0$

STEP 8

Now, we need to solve the quadratic equation for $n$ .
$n^2 + n - 342 = 0$

STEP 9

Factor the quadratic equation. We are looking for two numbers that multiply to -342 and add up to 1.

STEP 10

Find the factors of -342 that add up to 1. These factors are 18 and -19.

STEP 11

Write the factored form of the equation using the factors found in STEP_10.
$(n + 19)(n - 18) = 0$

STEP 12

Set each factor equal to zero and solve for $n$ .
$n + 19 = 0 \quad \text{or} \quad n - 18 = 0$

STEP 13

Solve the first equation for $n$ .
$n = -19$

STEP 14

Solve the second equation for $n$ .
$n = 18$

STEP 15

Now we have two possible values for $n$ , which are -19 and 18. We will use these to find the two pairs of consecutive integers.

STEP 16

Find the consecutive integer for $n = -19$ .
$n + 1 = -19 + 1 = -18$

STEP 17

The first pair of consecutive integers is -19 and -18.

STEP 18

Find the consecutive integer for $n = 18$ .
$n + 1 = 18 + 1 = 19$

STEP 19

The second pair of consecutive integers is 18 and 19.
The possible pairs of consecutive integers whose squares sum up to 685 are (-19, -18) and (18, 19).

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