find two consecutive even integers such that the square of the smaller decreased by 4 is equal to 4 times the larger

Answer 1

First of all, let´s name these two numbers as a and b, being "a" the smallest number and "b" the bigger one.

They are consecutive and even numbers, so b = a + 2. (Eq. 1)

Now, rewrite each statement using "a" and "b".

The square of the smaller: a^2.

Decreased by four: -4

Equal to: =

Four times the larger: 4b

Putting this together:

a^2 - 4 = 4b. (Eq 2)

Then, replace the letter "b" in Eq2 with its expression found in Eq 1 and solve.

That is: a^2 - 4 = 4*(a + 2) (Eq 3)

Solving:

a^2 - 4 = 4a + 8

a^2 -4 - 4a -8 = 0

a^2 - 4a - 12 = 0

Solving the trinomial:

(a - 6)*(a + 2) = 0

a1 = 6

a2 = -2

Since b = a + 2,

b1 = 6 +2 = 8

b2 = -2 + 2 = 0

Therefore, your sets of numbers are a1, b1 and a2, b2

That is:

Set 1: +6 and +8

Set 2: -2 and 0