Question

Let $\alpha, \beta$ be the roots of the equation $x^2 - x + p = 0$ and $\gamma, \delta$ be the roots of the equation $x^2 - 4x + q = 0$; $p, q \in \mathbb{Z}$. If $\alpha, \beta, \gamma, \delta$ are in G.P., then $|p+q|$ equals :

• A. 16
• B. 32
• C. 34
• D. 38

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The problem involves roots of quadratic equations forming a Geometric Progression (G.P.).
We need to use the relations between roots and coefficients for quadratic equations and express the roots in terms of a common ratio.
: Key Formula or Approach:
For a quadratic equation $ax^2 + bx + c = 0$, sum of roots $= -b/a$ and product of roots $= c/a$.
Let the roots in G.P. be $a, ar, ar^2, ar^3$.
Step 2: Detailed Explanation:
Given $x^2 - x + p = 0$ has roots $\alpha, \beta$.
Let $\alpha = a$ and $\beta = ar$.
Sum: $a + ar = 1 \implies a(1 + r) = 1 \quad \dots(i)$
Product: $a(ar) = p \implies a^2r = p \quad \dots(ii)$
Given $x^2 - 4x + q = 0$ has roots $\gamma, \delta$.
Let $\gamma = ar^2$ and $\delta = ar^3$.
Sum: $ar^2 + ar^3 = 4 \implies ar^2(1 + r) = 4 \quad \dots(iii)$
Product: $ar^2(ar^3) = q \implies a^2r^5 = q \quad \dots(iv)$
Dividing (iii) by (i):
\[ \frac{ar^2(1 + r)}{a(1 + r)} = \frac{4}{1} \implies r^2 = 4 \implies r = \pm 2 \]
If $r = 2$, then from (i), $a(3) = 1 \implies a = 1/3$.
Then $p = (1/3)^2(2) = 2/9$, but $p \in \mathbb{Z}$, so this is rejected.
If $r = -2$, then from (i), $a(1 - 2) = 1 \implies a = -1$.
Now, $p = a^2r = (-1)^2(-2) = -2$.
And $q = a^2r^5 = (-1)^2(-2)^5 = -32$.
Since $p, q \in \mathbb{Z}$, these values are valid.
$|p + q| = |-2 - 32| = |-34| = 34$.
Step 3: Final Answer:
The value of $|p+q|$ is 34.