Animated Solution for Mathematics - Quadratic Equations: Let $a\in \mathbb{R}$ and let $\alpha ,\beta$ be the roots of the equation $x^2+60^{\frac{1}{4}}x+a=0$. If ${\alpha }^4+{\beta }^4=-30$, then the product of all possible values of $a$ is

Imagine you are standing before a seemingly daunting algebraic challenge. You are given a quadratic equation, $x^2+60^{\frac{1}{4}}x+a=0$, and told that its roots, $\alpha$ and $\beta$, satisfy a mysterious condition: ${\alpha }^4+{\beta }^4=-30$.

At first glance, this looks like a problem that might require finding the roots themselves. But as an elite student, you know better. The secret to solving this lies not in finding the roots, but in understanding their collective behavior.

Whenever we deal with the roots of a polynomial, our first instinct should be to invoke Vieta's formulas. They are the bridge between the roots and the coefficients.

For our equation, the sum and product of the roots are:

This is our toolkit. We do not need to know what $\alpha$ or $\beta$ are individually; we only need their sum and product to unlock the higher powers.

We need to reach ${\alpha }^4+{\beta }^4$. We build it up, step by step, starting with the sum of squares:

Substituting our known values, we get ${\alpha }^2+{\beta }^2=(-60^{\frac{1}{4}}{)}^2-2a$. Simplifying this, the negative sign vanishes, and the power becomes $60^{\frac{2}{4}}$, which is simply $\sqrt{60}$.

Thus, we have:

Now, we take the next leap. We use the same identity structure for the fourth powers:

This is where the magic happens. We have successfully reduced a fourth-degree problem into a simple quadratic expression in terms of $a$.

Let us substitute our expression for the sum of squares into the identity:

Expanding the square of $(\sqrt{60}-2a)$, we obtain $60-4a\sqrt{60}+4a^2$. Subtracting $2a^2$, we get:

Simplifying $\sqrt{60}$ as $2\sqrt{15}$, the term $4a\sqrt{60}$ becomes $8a\sqrt{15}$. Our equation now stands as:

Bringing the $-30$ to the left, we arrive at the final quadratic:

The problem asks for the product of all possible values of $a$. The values of $a$ that satisfy the condition are the roots of the quadratic equation $2a^2-8\sqrt{15}a+90=0$.

We do not need to solve for $a$ using the quadratic formula. We simply use Vieta's formula for the product of the roots:

And there it is. The complexity dissolves into a simple, elegant result. By trusting the identities and avoiding the trap of calculating the roots directly, we have conquered the problem.

The final answer is 45.