PLEASE help me.. thank you in advance.

Question

Consider two consecutive positive integers such that the square of the second integer added to 2 times the first is equal to 78 .

Find the integers by solving the equation.

Answer 1

Let the first integer be x. Then , the next consecutive one is x+1.

The square of the second + 2 times the first is

(x+1)^2 + 2x

This equals 78,

Then, (x+1)^2 + 2x = 78

x^2 + 2x + 1 + 2x = 78

x^2 + 4x - 77 = 0

(x + 11)(x - 7) = 0

We are given that x is positive

x = 7 The next integer is 8

As a check,

2 * 7 + 8^2 = 14 + 64 = 78

Answer 2

Let x = the first number

x+1 the second number

The first number x is doubled and the second x +1 is squared. The sum of the two numbers is 78.

The equation would be 2x + (x+1)^2 = 78. We need to multiply (x + 1)(x+1) = x^2 + 2x +1. The whole equation is 2x + x^2 + 2x + 1= 78

terms and solve:

x^2 + 4x + 1 = 78

If we try to factor the left side as it is, the equation wont factor to integers. We need to complete the square with the following steps:

Subract 1 from both sides to make the right side.

Factor x^2 + 4x = 77

We can complete the square by taking half of the middle term, square it and add it to both sides. In this case we take (4* 1/2)^2 = 4

Add 4 to both sides and factor the equation into a perfect square.

x^2 + 4x + 4 = 77 + 4

(x+2)^2 = 81

x+2 = 9 x+2 = -9

take the square root of each side to get

x = 7, x + 2 = 9

x = -7 and x + 2 = -5

The integers must be positive so the final answer is 7 and 9