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3 votes
Is x+y=8

and x^2+y^2=80
then what is xy?

a. 16
b. -8
c. 8
d.-16
e.-32

1 Answer

4 votes

Answer: Given the two equations x + y = 8 and x^2 + y^2 = 80, we can use the first equation to solve for x or y and substitute that expression into the second equation.

Let's solve for y:

y = 8 - x

Now we can substitute this expression for y into the second equation:

x^2 + (8 - x)^2 = 80

Expanding the square and collecting terms gives:

x^2 + 64 - 16x + x^2 = 80

Combining like terms:

2x^2 - 16x + 16 = 80

Moving all terms to the left side:

2x^2 - 16x - 64 = 0

Using the quadratic formula, we can solve for x:

x = (-b ± √(b^2 - 4ac)) / (2a)

where a = 2, b = -16, and c = -64

x = (-(-16) ± √((-16)^2 - 4 * 2 * -64)) / (2 * 2)

x = (16 ± √(256 + 512)) / 4

x = (16 ± √768) / 4

x = (16 ± 8√6) / 4

Since x and y must be real numbers, we reject the negative square root. Therefore, x = (16 + 8√6) / 4.

Finally, substituting x into the first equation to find y:

y = 8 - x = 8 - (16 + 8√6) / 4 = (8 - 16) / 4 - 8√6 / 4 = -8 / 4 - 8√6 / 4 = -2 - 2√6.

Therefore, xy = x * y = (16 + 8√6) / 4 * (-2 - 2√6) = -16 - 16√6.

The answer is d. -16.

Explanation:

User Sujith Kumar KS
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