Let α, β be the roots of the equation x² - √2x + √6 = 0 and 1/α² + 1, 1/β² + 1 be the roots of the e

Q: Wait, why can't we just solve for $\alpha$ and $\beta$ using the quadratic formula? It seems more direct.

That is a classic trap! While you *could* technically solve for $\alpha$ and $\beta$ using the quadratic formula, you would be dealing with nested radicals and complex arithmetic that is prone to calculation errors. In JEE Advanced, the goal is to recognize the structure. By using Vieta's formulas, we bypass the need to know the roots individually and work directly with their properties. It is faster, cleaner, and significantly less error-prone.

Q: Why did we calculate $a+b$ as a single block instead of finding $a$ and $b$ separately?

Great observation! Look closely at the final equation we need to solve: $x^2 - (a + b - 2)x + (a + b + 2) = 0$. Notice that the coefficients only depend on the sum $(a + b)$. By calculating $(a + b)$ directly, we save time and reduce the chance of making a sign error while manipulating $a$ and $b$ individually. It is a strategic shortcut that separates the amateurs from the pros.

Q: How do I know if the roots are real or complex without actually solving the final quadratic?

You check the discriminant, $D = B^2 - 4AC$. In our final equation $6x^2 + 17x + 7 = 0$, we have $A=6, B=17, C=7$. Calculating $D = 17^2 - 4(6)(7) = 289 - 168 = 121$. Since $D > 0$ and is a perfect square, the roots are not only real but also rational. Always check the discriminant before jumping into the quadratic formula!

Question

Let $\alpha, \beta$ be the roots of the equation $x^2 - \sqrt{2}x + \sqrt{6} = 0$ and $\frac{1}{\alpha^2} + 1, \frac{1}{\beta^2} + 1$ be the roots of the equation $x^2 + ax + b = 0$. Then the roots of the equation $x^2 - (a + b - 2)x + (a + b + 2) = 0$ are :

Options:

(A) non-real complex numbers

(B) real and both negative

(C) real and both positive

(D) real and exactly one of them is positive

Sigma Percentile · Accepted Answer

### The Symphony of Symmetric Roots: A Masterclass in Quadratic Elegance

Imagine you are standing at the threshold of a complex algebraic puzzle. You see the equation $x^2 - \sqrt{2}x + \sqrt{6} = 0$, and your instinct might be to reach for the quadratic formula to find $\alpha$ and $\beta$.

Stop. Take a breath. In the world of JEE Advanced, the most elegant path is rarely the brute-force one.

This problem is a beautiful exercise in the power of symmetric functions. We are not interested in the individual values of $\alpha$ and $\beta$; we are interested in their collective behavior.

### Phase 1: The Foundation of Vieta

We begin by invoking the wisdom of Vieta. For any quadratic equation $Ax^2 + Bx + C = 0$, the sum of the roots is $-B/A$ and the product is $C/A$.

Applying this to our given equation $x^2 - \sqrt{2}x + \sqrt{6} = 0$, we immediately extract the gold:
$$\alpha + \beta = \sqrt{2}$$
$$\alpha\beta = \sqrt{6}$$

These two values are the keys to the entire kingdom. We do not need to know what $\alpha$ is; we only need to know how it dances with $\beta$.

### Phase 2: The Transformation

We are introduced to a second equation, $x^2 + ax + b = 0$, whose roots are $\frac{1}{\alpha^2} + 1$ and $\frac{1}{\beta^2} + 1$. We need to find the coefficients $a$ and $b$.

We know that for this new equation, the sum of the roots is $-a$:
$$-a = \left( \frac{1}{\alpha^2} + 1 \right) + \left( \frac{1}{\beta^2} + 1 \right)$$

By grouping the terms, we get:
$$-a = \left( \frac{\alpha^2 + \beta^2}{(\alpha\beta)^2} \right) + 2$$

Now, we calculate the numerator using the identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$:
$$\alpha^2 + \beta^2 = (\sqrt{2})^2 - 2\sqrt{6} = 2 - 2\sqrt{6}$$

Substituting this back into our expression for $-a$:
$$-a = \frac{2 - 2\sqrt{6}}{6} + 2 = \frac{1 - \sqrt{6}}{3} + 2 = \frac{7 - \sqrt{6}}{3}$$

Thus, we find:
$$a = \frac{\sqrt{6} - 7}{3}$$

### Phase 3: The Strategic Shortcut

Now, we calculate $b$, the product of the roots of the second equation:
$$b = \left( \frac{1}{\alpha^2} + 1 \right) \left( \frac{1}{\beta^2} + 1 \right) = \frac{1}{(\alpha\beta)^2} + \frac{\alpha^2 + \beta^2}{(\alpha\beta)^2} + 1$$

Substituting our known values:
$$b = \frac{1}{6} + \frac{2 - 2\sqrt{6}}{6} + 1 = \frac{1 + 2 - 2\sqrt{6} + 6}{6} = \frac{9 - 2\sqrt{6}}{6}$$

Now, adding $a$ and $b$ to simplify future calculations:
$$a + b = \frac{2\sqrt{6} - 14}{6} + \frac{9 - 2\sqrt{6}}{6} = -\frac{5}{6}$$

The beauty of this cancellation is why we love mathematics!

### Phase 4: The Final Reveal

With $a + b = -5/6$, our target equation becomes:
$$x^2 - \left( -\frac{5}{6} - 2 \right)x + \left( -\frac{5}{6} + 2 \right) = 0$$

Simplifying the coefficients:
$$x^2 - \left( -\frac{17}{6} \right)x + \frac{7}{6} = 0$$

Multiplying by 6, we arrive at the final quadratic equation:
**$$6x^2 + 17x + 7 = 0$$**

Solving this using the quadratic formula, we find the roots to be $-1/2$ and $-7/3$. We have successfully navigated the maze, not by brute force, but by understanding the underlying structure of the roots.

Select Answer:

Visualized Solution

Sum and Product of Roots

Calculate $α^{2} + β^{2}$

Sum of Roots for Second Equation

Solve for $a$

Product of Roots for Second Equation

Solve for $b$

Calculate $a + b$

Form the Final Equation

Find the Roots

Final Conclusion

The Sigma Insight: Relation Between Roots and Coefficients

The Symphony of Symmetric Roots

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

The numbers of pairs $(a, b)$ of real numbers, such that whenever $α$ is a root of the equation $x^{2} + a x + b = 0$ , $α^{2} - 2$ is also a root of this equation, is :

If 2 and 6 are the roots of the equation $a x^{2} + b x + 1 = 0$ , then the quadratic equation, whose roots are $\frac{1}{2 a + b}$ and $\frac{1}{6 a + b}$ , is :

Let $α, β$ be the roots of the equation $(x - a) (x - b) = c, c \neq = 0$ . Then the roots of the equation $(x - α) (x - β) + c = 0$ are

Let $- π /6 < θ < - π /12$ . Suppose $α_{1}$ and $β_{1}$ are the roots of the equation $x^{2} - 2 x sec θ + 1 = 0$ and $α_{2}$ and $β_{2}$ are the roots of the equation $x^{2} + 2 x tan θ - 1 = 0$ . If $α_{1} > β_{1}$ and $α_{2} > β_{2}$ , then $α_{1} + β_{2}$ equals

Let $α, β$ be the roots of the equation $x^{2} + 22 x - 1 = 0$ . The quadratic equation, whose roots are $α^{4} + β^{4}$ and $\frac{1}{10} (α^{6} + β^{6})$ , is :

If $α$ and $β$ ( $α < β$ ) are the roots of the equation $x^{2} + b x + c = 0$ , where $c < 0 < b$ , then

Let $p$ and $q$ be real numbers such that $p \neq = 0, p^{3} \neq = - q$ . If $α$ and $β$ are nonzero complex numbers satisfying $α + β = - p$ and $α^{3} + β^{3} = q$ , then a quadratic equation having $α / β$ and $β / α$ as its roots is

Difference between the corresponding roots of $x^{2} + a x + b = 0$ and $x^{2} + b x + a = 0$ is same and $a \neq = b$ , then

If $α$ and $β$ are the roots of the quadratic equation $x^{2} + x sin θ - 2 sin θ = 0, θ \in (0, π /2)$ , then $\frac{α ^{12} + β ^{12}}{( α ^{- 12} + β ^{- 12} ) ( α - β ) ^{24}}$ is equal to :

If $α, β$ , where $α < β$ , are the roots of the equation $λ x^{2} - (λ + 3) x + 3 = 0$ such that $\frac{1}{α} - \frac{1}{β} = \frac{1}{3}$ , then the sum of all possible values of $λ$ is

Select Answer:

Visualized Solution

Sum and Product of Roots

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α2+β2

Sum of Roots for Second Equation

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Product of Roots for Second Equation

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a+b