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SOURCE:COMPETITION Number of Problems: 11. FOR PRINT ::: (Book)
Real numbers and satisfy the equation . What is ?
If we complete the square after bringing the and terms to the other side, we get . Squares of real numbers are nonnegative, so we need both and to be . This obviously only happens when and .
For which of the following values of does the equation have no solution for ?
The domain over which we solve the equation is .
We can now cross-multiply to get rid of the fractions, we get .
Simplifying that, we get . Clearly for  we get the equation which is never true.
For other , one can solve for : , hence . We can easily verify that for none of the other four possible values of is this equal to or , hence there is a solution for in each of the other cases.
Using factoring:
or
So and are and .
Therefore the answer is
We can use the sum and product of a quadratic:
Two different positive numbers and each differ from their reciprocals by . What is ?
Each of the numbers and is a solution to .
Hence it is either a solution to , or to . Then it must be a solution either to , or to .
There are in total four such values of , namely .
Out of these, two are positive: and . We can easily check that both of them indeed have the required property, and their sum is .
The quadratic equation has roots twice those of , and none of and is zero. What is the value of ?
Let have roots and . Then
so and . Also, has roots and , so
and and . Thus .
Indeed, consider the quadratics .