Find two consecutive integers, the sum of whose squares is 1 more than twelve times the larger.
(Please provide the steps/explanation for solving this problem)
two consecutive integers: x and x+1
<the sum of whose squares is 1 more than twelve times the larger..>
x^2 + (x+1)^2 = 12(x+1) + 1
expand and solve as a quadratic equation.
(see if you can come up with 6,7 or -1,0 )
Thanks for the help. :)
Let's assume the two consecutive integers as x and (x+1). We can start by translating the given information into an equation.
According to the problem, the sum of their squares is 1 more than twelve times the larger integer. We can express this as:
x^2 + (x+1)^2 = 12(x+1) + 1
Now, let's solve this equation step by step to find the values of x and (x+1).
Expanding the equation:
x^2 + (x^2 + 2x + 1) = 12x + 12 + 1
Simplifying:
2x^2 + 2x + 1 = 12x + 13
Bringing all terms to one side:
2x^2 + 2x - 12x - 13 + 1 = 0
2x^2 - 10x - 12 = 0
Now, we can solve this quadratic equation by factoring or using the quadratic formula. Let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Here, a = 2, b = -10, and c = -12. Plugging these values in:
x = (-(-10) ± √((-10)^2 - 4(2)(-12))) / (2(2))
x = (10 ± √(100 + 96)) / 4
x = (10 ± √196) / 4
x = (10 ± 14) / 4
Now, we have two possible values for x:
1) When using the positive square root:
x = (10 + 14) / 4 = 24 / 4 = 6
2) When using the negative square root:
x = (10 - 14) / 4 = -4 / 4 = -1
So, the two consecutive integers can be either 6 and 7 or -1 and 0.