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Let's call the two numbers a and b. We know that ab = 280 and a + b = 27. Solving the second equation for b gives us b = 27 - a. We can then plug this equation into ab = 280 (since we know have b in terms of a). This gives us:
a(27 - a) = 280
Distribute and subtract 280 from both sides:
-a2 + 27 - 280 = 0
Use the quadratic formula to solve for a:
This step does not have any real solutions. There are no two real numbers that add to 280 and multiply to 27. You could use imaginary numbers. The quadratic formula shows that a could be about 13.5 - 9.887i. Plugging this value into one of the above equations (such as a + b = 27), and solving for b results in b equaling about 13.9 + 9.887i. (It could also come out so that a is about 13.9 + 9.887i and b is about 13.5 - 9.887i.)
So, there are no real solutions. If imaginary solutions are included, the two numbers could be about 13.5-9.887i and 13.5-9.887i.
(The exact forms are (1/2)(27 - isqr(391)) and (1/2)(27 + isqr(391)) The plus or minus in the quadratic formula takes care of what both numbers will be, so you don't have to substitute back if you find both possible values for a.)
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The numbers are: 55 and 55
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The two numbers that satisfy the given conditions are 4 and 1. When you add 4 and 1, you get 5. When you multiply 4 and 1, you get 4. The two numbers are 4 and 1.
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