Find two consecutive positive integers such that the square of the smaller increased by 4 times the larger is 64.

let the two numbers be x and x+1

x^2 + 4(x+1) = 64

x^2 + 4x + 4 - 64=0
x^2 + 4x - 60 = 0
(x+10)(x-6) = 0
x = -10 or x = 6
but you wanted only positive integers, so x = 6

the two numbers are 6 and 7

check:
What is 6^2 + 4(7) ??

Why did the chicken join a band? Because it had the perfect "beak" for playing the drums! Now, let's solve the problem at hand.

Let's assume the smaller positive integer is "x". Therefore, the larger positive integer would be "x + 1" since they are consecutive. We can set up the equation:

x^2 + 4(x + 1) = 64

Simplifying the equation, we have:

x^2 + 4x + 4 = 64

Moving things around, we get:

x^2 + 4x - 60 = 0

Using the quadratic formula, we find the value of x:

x = (-4 ± √(4^2 - 4 * 1 * (-60))) / 2 * 1

Simplifying, we get:

x = (-4 ± √(16 + 240)) / 2

x = (-4 ± √256) / 2

x = (-4 ± 16) / 2

x = (12 or -8) / 2

x = 6 or -4

Since we are looking for positive integers, we take x = 6 as our answer.

Therefore, the two consecutive positive integers are 6 and 7.

But hey, at least the chicken found its rhythm!

Let's call the smaller integer x, and the larger integer x+1, since they are consecutive positive integers.

According to the problem, the square of the smaller integer (x^2) increased by 4 times the larger integer (4(x+1)) is equal to 64 (x^2 + 4(x+1) = 64).

Now we can solve this equation step-by-step:

x^2 + 4(x+1) = 64

Expanding the equation:

x^2 + 4x + 4 = 64

Rearranging the terms:

x^2 + 4x + 4 - 64 = 0

Simplifying:

x^2 + 4x - 60 = 0

Now we can factorize this quadratic equation:

(x + 10)(x - 6) = 0

Setting each factor equal to zero and solving for x:

x + 10 = 0 OR x - 6 = 0

x = -10 OR x = 6

Since we are looking for positive integers, we discard the negative solution.

Therefore, the smaller integer is x = 6, and the larger integer is x+1 = 6+1 = 7.

So, the two consecutive positive integers that satisfy the given condition are 6 and 7.

To find two consecutive positive integers that satisfy the given condition, let's use algebraic equations.

Let's assume the smaller integer is x. The larger consecutive integer will be x + 1.

According to the given condition, the square of the smaller integer, increased by 4 times the larger integer, is equal to 64. We can write this as an equation:

x^2 + 4(x + 1) = 64

Expanding and simplifying the equation, we get:

x^2 + 4x + 4 = 64

Rearranging the equation, we have:

x^2 + 4x + 4 - 64 = 0

x^2 + 4x - 60 = 0

Now, we can solve this quadratic equation for x. We can either factor the equation or use the quadratic formula. Let's use factoring:

(x + 10)(x - 6) = 0

Setting each factor equal to zero and solving for x, we have:

x + 10 = 0 or x - 6 = 0

If we solve both equations, we find:

x = -10 or x = 6

Since we are looking for positive integers, we discard the negative solution (x = -10) and consider only the positive value x = 6.

So, the smaller integer is 6 and the larger consecutive integer is 6 + 1 = 7.

Therefore, the two consecutive positive integers that satisfy the given condition are 6 and 7.