Quadratic Equations and Geometric Progression (G.P.) Roots

We are given two quadratic equations: 1. $x^2 - x + p = 0$, with roots $\alpha, \beta$. 2. $x^2 - 4x + q = 0$, with roots $\gamma, \delta$. We are also given that $p, q \in \mathbf{Z}$ (integers) and that the roots $\alpha, \beta, \gamma, \delta$ are in Geometric Progression (G.P.). We need to find the value of $|p+q|$.

Applying Vieta's Formulas

From Vieta's formulas applied to the given equations:

• For the first equation: $\alpha + \beta = 1$ and $\alpha \beta = p$.
• For the second equation: $\gamma + \delta = 4$ and $\gamma \delta = q$.

Roots in Geometric Progression

Let the four roots in G.P. be denoted by $a, ar, ar^2, ar^3$, where $a$ is the first term and $r$ is the common ratio. We need to consider how these roots can be assigned to $\alpha, \beta, \gamma, \delta$.

Case 1: Grouping Roots

Assume the roots of the first equation are the first two terms of the G.P., and the roots of the second equation are the next two terms. Let $\{\alpha, \beta\} = \{a, ar\}$ and $\{\gamma, \delta\} = \{ar^2, ar^3\}$.

• From $\alpha + \beta = 1$: $a + ar = 1 \implies a(1+r) = 1$.
• From $\alpha \beta = p$: $a(ar) = p \implies a^2r = p$.
• From $\gamma + \delta = 4$: $ar^2 + ar^3 = 4 \implies ar^2(1+r) = 4$.
• From $\gamma \delta = q$: $(ar^2)(ar^3) = q \implies a^2r^5 = q$.

Solving for Common Ratio $r$

Divide the equation $ar^2(1+r) = 4$ by $a(1+r) = 1$: $ \frac{ar^2(1+r)}{a(1+r)} = \frac{4}{1} $ $ r^2 = 4 $ This gives $r = 2$ or $r = -2$. We must check which value yields integer values for $p$ and $q$.

Evaluating Case $r=2$

If $r=2$, substitute into $a(1+r) = 1$: $ a(1+2) = 1 \implies 3a = 1 \implies a = \frac{1}{3} $ Now find $p$: $ p = a^2r = \left(\frac{1}{3}\right)^2 (2) = \frac{1}{9} \times 2 = \frac{2}{9} $ Since $p$ must be an integer ($p \in \mathbf{Z}$), $r=2$ is not a valid solution.

Evaluating Case $r=-2$

If $r=-2$, substitute into $a(1+r) = 1$: $ a(1+(-2)) = 1 \implies a(-1) = 1 \implies a = -1 $ Now find $p$: $ p = a^2r = (-1)^2 (-2) = (1)(-2) = -2 $ Since $p = -2$ is an integer, this is a valid value. Now find $q$: $ q = a^2r^5 = (-1)^2 (-2)^5 = (1)(-32) = -32 $ Since $q = -32$ is an integer, this is also a valid value. The roots are $\alpha = -1$, $\beta = (-1)(-2) = 2$; $\gamma = (-1)(-2)^2 = -4$, $\delta = (-1)(-2)^3 = 8$. These are $-1, 2, -4, 8$, which form a G.P. with $r=-2$. The quadratic equations are $x^2 - x - 2 = 0$ and $x^2 - 4x - 32 = 0$, confirming $p=-2$ and $q=-32$.

Calculating $|p+q|$

Using the integer values found: $p=-2$ and $q=-32$. $ |p+q| = |-2 + (-32)| = |-34| = 34 $

Other possible groupings of roots between the two equations lead to non-integer values for $p$ or $q$. Therefore, the only valid case yields $|p+q|=34$.