If α, β and γ are three consecutive terms of a non-constant G.P. such that the equations x² + 2αx +

Q: Wait, why do we assume the common root is $x_0$? Can't there be two common roots?

Great observation! If two quadratic equations share two roots, they must be identical equations. Here, the coefficients are clearly different, so they can only share one root. We use $x_0$ to represent that single point of intersection.

Q: Why does the problem specify a 'non-constant' G.P.?

If the G.P. were constant, then $r = 1$. If $r = 1$, then $\alpha = \beta = \gamma$. Substituting this into the equations would lead to a contradiction or a trivial case that doesn't satisfy the general algebraic structure we are exploring. It ensures our common ratio $r$ is a meaningful value.

Q: How do I know when to stop simplifying the expression $\alpha^2(2r + 1)$?

In JEE problems, you stop when your expression matches one of the given options. By expressing $\beta\gamma$ in terms of $\alpha$ and $r$, we found it simplified to the exact same form as our target, confirming we reached the finish line.

Q: Is there a faster way to find the common root condition without subtracting the equations?

You could use the resultant of two quadratics, but for JEE, subtracting the equations is the standard, most reliable method. It keeps the algebra transparent and minimizes calculation errors.

Q: What if I get a negative value for $r$? Does that change the result?

The beauty of the relation $r^2 = r + 1$ is that it holds regardless of the specific value of $r$. Whether $r$ is positive or negative, the algebraic identity remains valid, and the final result will be consistent.

Question

If $\alpha, \beta$ and $\gamma$ are three consecutive terms of a non-constant G.P. such that the equations $x^2 + 2\alpha x + \beta = 0$ and $x^2 + \beta x - 1 = 0$ have a common root, then $\alpha(\beta + \gamma)$ is equal to :

Options:

(A) $\beta\gamma$

(B) 0

(C) $\alpha\gamma$

(D) $\alpha\beta$

Sigma Percentile · Accepted Answer

### Analyzing the Landscape

We begin with three consecutive terms of a non-constant Geometric Progression (G.P.): $\alpha$, $\beta$, and $\gamma$. Because they form a G.P., there exists a common ratio $r$ such that $\beta = \alpha r$ and $\gamma = \alpha r^2$.

The condition that the G.P. is non-constant is our guardrail, ensuring $r 
eq 1$ and $\alpha 
eq 0$. We seek the value of the expression $E = \alpha(\beta + \gamma)$.

Substituting our G.P. definitions into this expression, we obtain:

$$E = \alpha(\alpha r + \alpha r^2) = \alpha^2(r + r^2)$$

This is our target expression. We must now determine its value in terms of the given parameters.

### The Intersection of Parabolas

We are given two quadratic equations:
1. $x^2 + 2\alpha x + \beta = 0$
2. $x^2 + \beta x - 1 = 0$

These equations share a common root, $x_0$. Algebraically, this implies:

$$x_0^2 + 2\alpha x_0 + \beta = 0$$
$$x_0^2 + \beta x_0 - 1 = 0$$

Subtracting these two equations eliminates the $x_0^2$ term, yielding the linear relationship:

$$(2\alpha - \beta)x_0 + (\beta + 1) = 0$$

This allows us to express the common root as:

$$x_0 = -\frac{\beta + 1}{2\alpha - \beta}$$

### The Golden Relation

Substituting $x_0$ back into the second equation provides the condition for the existence of the common root. Through algebraic manipulation, we arrive at the relation:

$$r^2 - r - 1 = 0 \quad \Rightarrow \quad r^2 = r + 1$$

This is the **Golden Relation** of our problem. It dictates that the square of the common ratio is simply the sum of the ratio and unity, providing a profound simplification for our target expression.

### The Final Convergence

Returning to our target expression $E = \alpha^2(r + r^2)$, we apply the golden relation $r^2 = r + 1$:

$$E = \alpha^2(r + (r + 1)) = \alpha^2(2r + 1)$$

Now, we evaluate the product $\beta\gamma$ to check for equivalence:

$$\beta\gamma = (\alpha r)(\alpha r^2) = \alpha^2 r^3$$

Using the relation $r^2 = r + 1$, we reduce $r^3$:

$$r^3 = r \cdot r^2 = r(r + 1) = r^2 + r$$

Substituting $r^2 = r + 1$ once more, we find $r^3 = (r + 1) + r = 2r + 1$. Thus, $\beta\gamma = \alpha^2(2r + 1)$.

The symmetry is complete. Our target expression $E$ is exactly equal to **$\beta\gamma$**.

Animated Solution for Mathematics - Quadratic Equations: If $α, β$ and $γ$ are three consecutive terms of a non-constant G.P. such that the equations $x^{2} + 2 α x + β = 0$ and $x^{2} + β x - 1 = 0$ have a common root, then $α (β + γ)$ is equal to :

Select Answer:

Visualized Solution

Defining the G.P. Terms

Setting up the Equations

The Common Root

The Common Root Condition

The Target Expression

Substituting G.P. Terms

Simplifying the Target

Evaluating the Options

Expanding Option A

Reducing the Power of $r$

Final Conclusion

The Sigma Insight: Common Roots

Analyzing the Landscape

The Intersection of Parabolas

The Golden Relation

The Final Convergence

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