Question

10. The larger of two positive integers is three more than the smaller. If twice the

square of the smaller integer is increased by three times the larger integer, the
result is 74. Find the integers.

Answer 1

Correct answer: n₁ = 8 and n₂ = 5

Explanation:

If i understand correctly you claim that the first number by three is greater than the second.

We will form a system of equations according to the given data

Let be first positive integer n₁ and second n₂

n₁ = n₂ + 3

2 n₂² + 3 n₁ = 74

If we replace the first number n₁ = n₂ + 3 in the second equation, we get:

2 n₂² + 3 (n₂ + 3) = 74 ⇒ 2 n₂² + 3 n₂ + 9 = 74 ⇒

2 n₂² + 3 n₂ - 65 = 0

n₁₂ = (-b ±√b² - 4 a c) / 2 a

n₁₂ = (-3 ±√3² - 4 · 2 · (-65)) / 2 · 2 = (-3 ±√9 + 520) / 4

n₁₂ = (-3 ±√529) / 4 = (-3 ± 23) / 4

We accept the positive solution

n₂ = (-3 + 23) / 4 = 20 / 4 = 5

n₂ = 5

n₁ = 5 + 3 = 8

n₁ = 8

God is with you!!!

Answer 2

Answer: The integers are 8 and 5

Explanation:

An integer is a whole number.

Let x represent the larger positive integer.

Let y represent the smaller positive integer.

The larger of two positive integers is three more than the smaller. It means that

x = y + 3

If twice the square of the smaller integer is increased by three times the larger integer, the result is 74. It means that

2y²+ 3x = 74- - - - - - - - - - -1

Substituting x = y + 3 into equation 1, it becomes

2y² + 3(y + 3) = 74

2y² + 3y + 9 = 74

2y² + 3y + 9 - 74 = 0

2y² + 3y - 65 = 0

2y² + 13y - 10y - 65 = 0

y(2y + 13) - 5(2y + 13) = 0

y - 5 = 0 or 2y - 13 = 0

y = 5 or y = 13/2

Since an integer is a whole number, then

y = 5

x = y + 3 = 5 + 3

x = 8