I know that -b/a is the sum of quadratic solutions and c/a is the product of solutions and I wonder if that is the key to solving this problem. Other than that I HAVE NO IDEA how to even begin. The solution, according to my instructions, must be arrived at by the quadratic formula.
Here is the problem:
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Yes, it is true that the solutions to a quadratic equation are
\(\displaystyle ax^2 + bx + c \implies x = \dfrac{-\ b + \sqrt{b^2 - 4ac}}{2a} \text { or } x = \dfrac{- \ b - \sqrt{b^2 - 4ac}}{2a}.\)
That formula comes up often enough that it is convenient to remember it.
And from that formula, one can easily deduce that the sum of the solutions is
\(\displaystyle \dfrac{-\ b + \sqrt{b^2 - 4ac}}{2a} + \dfrac{-\ b - \sqrt{b^2 - 4ac}}{2a} =\)
\(\displaystyle \dfrac{-\ b - b + \sqrt{b^2 - 4ac} - \sqrt{b^2 - 4ac}}{2a} = \dfrac{-\ 2b}{2a} = -\ \dfrac{b}{a}.\)
And deduce that the product of the solutions is
\(\displaystyle \dfrac{-\ b + \sqrt{b^2 - 4ac}}{2a} * \dfrac{-\ b - \sqrt{b^2 - 4ac}}{2a} =\)
\(\displaystyle \dfrac{(-\ b)^2 - b\{\sqrt{b^2 - 4ac} - \sqrt{b^2 - 4ac}\} - \{\sqrt{b^2 - 4ac}\}^2}{2a * 2a} = \)
\(\displaystyle \dfrac{b^2 - b(0) - (b^2 - 4ac)}{4a^2} = \dfrac{4ac}{4a^2} = \dfrac{c}{a}.\)
But those two facts are not worth memorizing.
Math is not a bunch of facts to be memorized. It is a small set of techniques that you learn how to apply in different situations. You memorized some facts and wondered if two of those facts might be helpful in solving a specific problem. That is not a useful approach. There are literally an infinite number of facts about the real numbers.
The fundamental idea in algebra is that you find unknown numbers by finding what they are equal to. This problem is amazingly easy if you remember that fundamental idea because the equalities are laid on your plate.
\(\displaystyle x + y = 2 \text { and } xy = -\ 1.\)
You have two equations in two unknowns. There is a general technique for solving such systems of equations called substitution.
\(\displaystyle y = 2 - x \implies -\ 1 = xy = x(2 - x) = 2x - x^2 \implies x^2 - 2x - 1 = 0.\)
Now, as I said, such quadratic equations in one unknown come up often enough that it is convenient to memorize a formula for solving them, but the formula is not necessary. There is a general technique for solving these equations called completing the square.
\(\displaystyle x^2 - 2x - 1 = 0 \implies x^2 - 2x - 1 + 2 = 2 \implies x^2 - 2x + 1 = 2 \implies\)
\(\displaystyle (x - 1)^2 = 2 \implies (x - 1) = \pm \sqrt{2} \implies x = 1 + \sqrt{2} \text { or } x = 1 - \sqrt{2} \implies\)
\(\displaystyle y = 2 - (1 + \sqrt{2}) = 1 - \sqrt{2} \text { or } y = 2 - (1 - \sqrt{2}) = 1 + \sqrt{2}.\)
Algebra is about applying a very small number of ideas and techniques to a very wide range of problems. It is not about amassing a huge number of mathematical facts to apply to the small proportion of problems to which those facts are relevant.
