Q:

Consider two consecutive positive integers such that the square of the second integer added to 3 times the first is equal to 85

Accepted Solution

A:
Answer:The Two Consecutive integers are 7 and 8.Step-by-step explanation:Let the first Integer be xSince the two numbers are consecutivetherefore the second number will be x+1Now given:the square of the second integer added to 3 times the first is equal to 85Hence framing the above sentence in mathematical form we get;$$(x+1)^2+3x=85$$Now Solving the above equation we get;since $$(a+b)^2 = a^2+2ab+b^2$$$$x^2+2x+1+3x=85$$Using addition property we get;$$x^2+5x+1-85=0$$Using Subtraction property we get;$$x^2+5x-84=0$$Now factorizing above equation we get;$$x^2+12x-7x-84=0\\x(x+12)-7(x+12)=0\\(x-7)(x+12)=0\\x-7=0 \ \ \ \ \ \ \ \ or \ \ \ \ \ \ x+12 = 0\\x= 7\ \ \ \ \ \ \ \ or \ \ \ \ \ \ \ \ x=-12$$Now we get 2 values of x which is 7 and -12.Since it is given that number is positive Integer hence the number will be 7and the other number would 7 + 1 = 8.