Question

Determine exactly all values of a for which y=(a^2 +1)(x^2 - 1) - ax +7 has two real zeros whose sum equals their product.

Answer 1

Hint:
1. Expand right hand side to get a polynomial in x.
\(~~~~f(x)=y=a^2x^2+x^2-ax-a^2+6\)
2. Consider the quadratic in x, solve for x using the quadratic formula:
\(A=a^2+1,~B=-a,~ C=-a^2+6\)
call the solutions x1 and x2.
x1 and x2 are functions of "a", i.e. are expressions in terms of "a".
3. Solve for "a" in the equation x1*x2=x1+x2. Reject complex values of a since they will make a complex coefficient in f(x), and hence result in complex roots of f(x).
4. substitute real roots of "a" into solutions x1 and x2 to make sure both x1 and x2 are real. Reject value(s) of "a" that make x1 or x2 complex.
5. The two remaining values of a are the required solution.
6. Substitute these value of "a" into the original equation and ensure that all conditions are satisfied, i.e. x1*x2=x1+x2. These values of "a" turn out to be negative integers.

Answer 2

* negative and positive integers.