Question

quadratic equations. A positive real number is 4 less than another. When 8 times the larger is added to the square of the smaller, the result is 48.

Answer 1

To solve the problem, we can use algebraic equations. Let x be the smaller real number and y be the larger real number. We know that y = x + 4 and that 8y + x^2 = 48.

Substituting y = x + 4 into the second equation, we get:

8(x + 4) + x^2 = 48

Expanding and rearranging, we get:

x^2 + 8x - 16 = 0

Using the quadratic formula, we can solve for x:

x = (-b ± sqrt(b^2 - 4ac)) / 2a

where a = 1, b = 8, and c = -16. Plugging in these values, we get:

x = (-8 ± sqrt(8^2 - 4(1)(-16))) / 2(1)

Simplifying, we get:

x = (-8 ± sqrt(96)) / 2

x = (-8 ± 4sqrt(6)) / 2

x = -4 ± 2sqrt(6)

Since x is a positive real number, we take the positive root:

x = -4 + 2sqrt(6)

Now that we have found x, we can use y = x + 4 to find y:

y = -4 + 2sqrt(6) + 4

y = 2 + 2sqrt(6)

Therefore, the two real numbers are approximately -4 + 2sqrt(6) and 2 + 2sqrt(6).