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

We have three consecutive terms of a non-constant geometric progression (G.P.) written as $$\alpha,\,\beta,\,\gamma.$$

If the common ratio of this G.P. is denoted by $$q,$$ then by definition of a G.P. we can write

$$\beta=\alpha q,\qquad \gamma=\beta q=\alpha q^2.$$

The first quadratic given in the statement is

$$\alpha x^2+2\beta x+\gamma=0.$$

Substituting $$\beta=\alpha q$$ and $$\gamma=\alpha q^2$$ into it we obtain

$$\alpha x^2+2(\alpha q)x+\alpha q^2=0.$$

Dividing every term by the non-zero number $$\alpha$$ gives the simpler form

$$x^2+2q\,x+q^2=0. \quad -(1)$$

The second quadratic supplied in the question is

$$x^2+x-1=0. \quad -(2)$$

We are told that equations (1) and (2) share a common root. Let this common root be denoted by $$k.$$ Hence

$$k^2+2qk+q^2=0 \qquad\text{and}\qquad k^2+k-1=0.$$

Subtracting the second equation from the first eliminates the $$k^2$$ terms:

$$\bigl(k^2+2qk+q^2\bigr)-\bigl(k^2+k-1\bigr)=0.$$

Simplifying, we get

$$2qk-k+q^2+1=0.$$

Taking $$k$$ common from the first two terms we find

$$k(2q-1)+q^2+1=0,$$

so that

$$k=-\dfrac{q^2+1}{\,2q-1\,}. \quad -(3)$$

The value of $$k$$ in (3) must also satisfy equation (2), namely $$k^2+k-1=0.$$ Substituting (3) in this quadratic therefore yields

$$\left(-\dfrac{q^2+1}{\,2q-1\,}\right)^{\!2}+\left(-\dfrac{q^2+1}{\,2q-1\,}\right)-1=0.$$

Multiplying through by $$(2q-1)^2$$ to clear denominators gives

$$(q^2+1)^2-(q^2+1)(2q-1)-(2q-1)^2=0.$$

Expanding every expression completely:

$$\bigl(q^4+2q^2+1\bigr)-\bigl(2q^3-q^2+2q-1\bigr)-\bigl(4q^2-4q+1\bigr)=0.$$

Combining like terms step by step:

$$q^4+2q^2+1-2q^3+q^2-2q+1-4q^2+4q-1=0,$$

which simplifies to

$$q^4-2q^3-q^2+2q+1=0.$$

On careful inspection this quartic can be recognised as a perfect square:

$$q^4-2q^3-q^2+2q+1=\bigl(q^2-q-1\bigr)^2.$$

Hence we must have

$$\bigl(q^2-q-1\bigr)^2=0\quad\Longrightarrow\quad q^2-q-1=0.$$

Because the progression is non-constant we discard the irrelevant possibility $$q=1,$$ retaining only the genuine relation

$$q^2=q+1. \quad -(4)$$

Our task is now to evaluate $$\alpha\bigl(\beta+\gamma\bigr).$$ Using $$\beta=\alpha q$$ and $$\gamma=\alpha q^2$$ we have

$$\beta+\gamma=\alpha q+\alpha q^2=\alpha q(1+q).$$

Therefore

$$\alpha(\beta+\gamma)=\alpha\bigl[\alpha q(1+q)\bigr]=\alpha^2q(1+q).$$

Next, let us compute $$\beta\gamma$$ in order to compare it with the above value:

$$\beta\gamma=(\alpha q)(\alpha q^2)=\alpha^2q^3.$$

We now invoke the crucial identity (4). From $$q^2=q+1$$ we directly obtain

$$q^3=q\cdot q^2=q(q+1)=q^2+q=(q+1)+q=2q+1.$$

On the other hand, the factor appearing in $$\alpha(\beta+\gamma)$$ is

$$q(1+q)=q+q^2=q+(q+1)=2q+1.$$

Thus we observe that

$$q(1+q)=q^3.$$

Multiplying both sides by $$\alpha^2$$ we get

$$\alpha^2q(1+q)=\alpha^2q^3,$$

which precisely means

$$\alpha(\beta+\gamma)=\beta\gamma.$$

Therefore $$\alpha(\beta+\gamma)$$ equals $$\beta\gamma,$$ matching Option A.

Hence, the correct answer is Option A.