Question

If \(\alpha, \beta\) are the roots of the equation \(x^{2}-px+q=0\) and \(\alpha>0\), \(\beta>0\), then \[ \alpha^{\frac{1}{4}}+\beta^{\frac{1}{4}}=(p+6\sqrt{q}+4q^{\frac{1}{4}}\sqrt{p+2\sqrt{q}})^{\kappa}, \] where \(\kappa\) is:

• A. $\frac{3}{2}$
• B. $2$
• C. $\frac{1}{3}$
• D. $\frac{1}{4}$

Hint:

When resolving complicated radical identities containing unknown exponents, try substituting simple numbers! If we let $\alpha = 1$ and $\beta = 1$, then $p = 2$ and $q = 1$. The left side becomes $1+1=2$, while the inside right expression simplifies to $2 + 6(1) + 4(1)\sqrt{2+2} = 8 + 4(2) = 16$. Solving $2 = (16)^K$ immediately yields $K = \frac{1}{4}$!

Answer 1

The Correct Option is D

Concept: For a standard quadratic equation $Ax^2 + Bx + C = 0$, Vieta's formulas map the roots directly to the coefficients via symmetric functions. We can construct radical expressions by progressively squaring symmetric root combinations. Step 1: Establish sum and product relations.
For the given quadratic equation $x^2 - px + q = 0$, Vieta's relations yield: $$\alpha + \beta = p \quad \text{and} \quad \alpha\beta = q$$ This implies the base operational radicals can be written as: $$\sqrt{\alpha\beta} = \sqrt{q} \quad \text{and} \quad (\alpha\beta)^{1/4} = q^{1/4}$$

Step 2: Square the target expression radical root.
Let our target radical sum expression be defined as $X = \alpha^{1/4} + \beta^{1/4}$. Let's compute its square: $$X^2 = \left(\alpha^{1/4} + \beta^{1/4}\right)^2 = \sqrt{\alpha} + \sqrt{\beta} + 2(\alpha\beta)^{1/4} = \sqrt{\alpha} + \sqrt{\beta} + 2q^{1/4}$$

Step 3: Determine the value of the nested square root sum.
To resolve the remaining term $(\sqrt{\alpha} + \sqrt{\beta})$, let us evaluate its square using our known symmetric sum components: $$\left(\sqrt{\alpha} + \sqrt{\beta}\right)^2 = \alpha + \beta + 2\sqrt{\alpha\beta} = p + 2\sqrt{q}$$ Taking the square root since $\alpha, \beta > 0$: $$\sqrt{\alpha} + \sqrt{\beta} = \sqrt{p + 2\sqrt{q}}$$ Substitute this evaluated component directly back into our expression for $X^2$: $$X^2 = \sqrt{p + 2\sqrt{q}} + 2q^{1/4}$$

Step 4: Square $X^2$ to construct the target polynomial bracket form.
Now, let us find the value of $X^4$ by expanding the square of our $X^2$ expression: $$X^4 = \left(\sqrt{p + 2\sqrt{q}} + 2q^{1/4}\right)^2 = \left(p + 2\sqrt{q}\right) + 4\sqrt{q} + 4q^{1/4}\sqrt{p + 2\sqrt{q}}$$ Combine the like terms containing $\sqrt{q}$: $$X^4 = p + 6\sqrt{q} + 4q^{1/4}\sqrt{p + 2\sqrt{q}}$$

Step 5: Equate components to find K.
Taking the fourth root of both sides gives: $$X = \left(p + 6\sqrt{q} + 4q^{1/4}\sqrt{p + 2\sqrt{q}}\right)^{1/4}$$ Comparing this structural extraction with the identity statement format $X = (\dots)^K$, we find: $$K = \frac{1}{4}$$