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Step-by-Step Solution
Step 1: Express the sum and product of roots for each quadratic
Given the quadratic equation
$x^2 - 3x + p = 0$ has roots $\\alpha$ and $\\beta$.
By Viète's formulas, we have:
$\\alpha + \\beta = 3$
$\\alpha\\beta = p$
Similarly, the roots of
$x^2 - 6x + q = 0$ are $\\gamma$ and $\\delta$. Hence:
$\\gamma + \\delta = 6$
$\\gamma\\delta = q$
Step 2: Set up the geometric progression (GP)
We know $\\alpha, \\beta, \\gamma, \\delta$ form a geometric progression. Let the first term of this GP be $a$ and the common ratio be $r$. Then:
$\\alpha = a$
$\\beta = ar$
$\\gamma = ar^2$
$\\delta = ar^3$
Step 3: Use sum of roots in GP
From $\\alpha + \\beta = 3$, we have:
$a + ar = 3$
$a(1 + r) = 3 \quad (1)$
From $\\gamma + \\delta = 6$, we have:
$ar^2 + ar^3 = 6$
$ar^2 (1 + r) = 6 \quad (2)$
Step 4: Determine the common ratio $r$
Divide equation (2) by (1):
$\\displaystyle \\frac{ar^2 (1 + r)}{a (1 + r)} = \\frac{6}{3}$
This simplifies to:
$r^2 = 2$, hence $r = \\sqrt{2}.$
Step 5: Determine $a$ using $a(1 + r) = 3$
Substitute $r = \\sqrt{2}$ into equation (1):
$a(1 + \\sqrt{2}) = 3$
$\\displaystyle a = \\frac{3}{1 + \\sqrt{2}}.$
Step 6: Compute $p = \\alpha\\beta$
Since $\\alpha = a$ and $\\beta = ar$, we have:
$\\alpha\\beta = a \\times (ar) = a^2 r.$
Substituting $a = \\frac{3}{1 + \\sqrt{2}}$ and $r = \\sqrt{2}$:
$\\displaystyle p = \\left(\\frac{3}{1 + \\sqrt{2}}\\right)^2 \\times \\sqrt{2}.$
For brevity, one can keep it as:
$\\displaystyle p = \\frac{9\\sqrt{2}}{(1 + \\sqrt{2})^2}.$
Step 7: Compute $q = \\gamma\\delta$
Since $\\gamma = ar^2$ and $\\delta = ar^3$, we have:
$\\gamma\\delta = ar^2 \\times (ar^3) = a^2 r^5.$
But $r^5 = (\\sqrt{2})^5 = 2^{5/2} = 2^2 \\times \\sqrt{2} = 4\\sqrt{2}.$ Hence:
$\\gamma\\delta = a^2 (4\\sqrt{2}).$
Substituting $a = \\frac{3}{1 + \\sqrt{2}}$:
$\\displaystyle q = \\left(\\frac{3}{1 + \\sqrt{2}}\\right)^2 \\times 4\\sqrt{2}
= \\frac{36\\sqrt{2}}{(1 + \\sqrt{2})^2}.$
Step 8: Find the required ratio $\\frac{2q + p}{2q - p}$
First compute $2q$ and then add/subtract $p$:
$\\displaystyle 2q = 2 \\times \\frac{36\\sqrt{2}}{(1 + \\sqrt{2})^2}
= \\frac{72\\sqrt{2}}{(1 + \\sqrt{2})^2}.$
So:
$
2q + p
= \\frac{72\\sqrt{2}}{(1 + \\sqrt{2})^2}
+ \\frac{9\\sqrt{2}}{(1 + \\sqrt{2})^2}
= \\frac{81\\sqrt{2}}{(1 + \\sqrt{2})^2},
$
$
2q - p
= \\frac{72\\sqrt{2}}{(1 + \\sqrt{2})^2}
- \\frac{9\\sqrt{2}}{(1 + \\sqrt{2})^2}
= \\frac{63\\sqrt{2}}{(1 + \\sqrt{2})^2}.
$
Therefore,
$\\displaystyle \\frac{2q + p}{2q - p}
= \\frac{\\frac{81\\sqrt{2}}{(1 + \\sqrt{2})^2}}{\\frac{63\\sqrt{2}}{(1 + \\sqrt{2})^2}}
= \\frac{81}{63}
= \\frac{9}{7}.$
Final Answer
The ratio $(2q + p) : (2q - p)$ is $9 : 7.$
