Find two consecutive positive integers such that the square of the second is added to 6 times the first is equal to 49?????
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trouble setting it up?
what should the formula look like?
well, there are two numbers to look for...
so two variables... which means you need two equations to solve it
Not the answer you are looking for? Search for more explanations.
let X be the first number and Y be the second number, can you dig an equation out of that sentence?
X and Y are consecutive integers. What does that mean in terms of X and Y? X = what?
Let x be a positive integer. Then, x +1 is the next consecutive positive integer
6x + (x+1)^2 = 49
6x + x^2 + 2x + 1 = 49
x^2 +8x -48 =0
(x +12)(x- 4)=0
Zero Product Property
x = -12 and x = 4
Integers must be positive so -12 is not considered.
Answer: 4 and 5