FULL WORKING WITH EXPLANATION

Question

Three consecutive integers are selected. The square of the middle one is subtracted from the sum of the squares of the other two, to give a result of 38. Find the smallest number.

Answer 1

Let the 1st integer = x

Let the 2nd integer = x + 1

Let the 3rd integer = x + 2

 

The 1st is the smallest, the 2nd is the middle, and the 3rd is the largest.  Using these variables, we can create equations.

 

[x² + (x + 2)²] - (x + 1)² = 38

 

 

Solve for x from this equation.

 

x² + (x + 2)(x + 2) - (x + 1)(x + 1) = 38

 

x² + (x² + 4x + 4) - (x² + 2x + 1) = 38

 

 

Combine like terms on the left side of equation.

 

x² + 2x + 3 = 38

 

 

Subtract 38 on both sides of the equation.

 

x² + 2x - 35 = 0

 

 

Factor using FOIL.

 

(x + 7)(x - 5) = 0

 

 

x = -7            or            x = 5

 

 

Now we verify to see if the result is 38 when plugging in the numbers.

 

5² + (5 + 2)² - (5 + 1)² = 38

 

25 + 49 - 36 = 38

 

38 = 38         true

 

 

(-7)² + (-7 + 2)² - (-7 + 1)² = 38

 

49 + 25 - 36 = 38

 

38 = 38          true

 

Both solutions satisfy, but the only smallest between the two is     x=-7

Answer 2

Let the three consecutive integers be a, a+1, and a+2

 

Then

 

a² + (a+2)² - (a+1)² = 38

 

Expand and collect terms.

 

a² + (a² + 4a + 4) - (a² + 2a + 1) = 38

 

a² + 2a + 3 = 35

 

a² + 2a - 35 = 0

 

Factor the expression to obtain a (or use the quadratic formula).

 

(a + 7)(a - 5) = 0

 

a = -7 or 5.  The smallest solution is a = -7.  The quadratic formula gives the same results.