Question

One positive number is 8 more than another. The sum of their squares is 130. Find both numbers

Answer 1

The two numbers are 3 and 11.

Step-by-step explanation:

Let's assume that the first number is x. Since the given number is 8 more than the other, the second number can be represented as x + 8. Now, we can set up an equation based on the sum of their squares:

x^2 + (x + 8)^2 = 130

Simplifying the equation, we get:

x^2 + x^2 + 16x + 64 = 130

Combining like terms, we have:

2x^2 + 16x + 64 = 130

Next, let's subtract 130 from both sides to get a quadratic equation:

2x^2 + 16x + 64 - 130 = 0

2x^2 + 16x - 66 = 0

Now, we can solve this quadratic equation using factoring, completing the square, or the quadratic formula. In this case, let's use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

Substituting the values a = 2, b = 16, and c = -66 into the formula, we obtain:

x = (-16 ± √(16^2 - 4(2)(-66))) / (2(2))

Simplifying further, we have:

x = (-16 ± √(256 + 528)) / 4

x = (-16 ± √(784)) / 4

x = (-16 ± 28) / 4

Therefore, we have two possible solutions for x:

x = (-16 + 28) / 4 = 12 / 4 = 3

x = (-16 - 28) / 4 = -44 / 4 = -11

So, the two numbers are 3 and 11.

Answer 2

first number: x

second number: x + 8

(x)² + (x + 8)² = 130

x² + x² + 16x + 64 = 130

2x² + 16x + 64 = 130

x² + 8x + 32 = 65 divided both sides by 2

x² + 8x - 33 = 0 subtracted 65 from both sides

(x + 11)(x - 3) = 0

x + 11 = 0 x - 3 = 0

x = -11 x = 3 disregard -11 because it's not a positive number

second number: x + 8 = (3) + 8 = 11

Answer: 3 and 11