Let be the roots of x 2 - x + p = 0 and be the roots of x 2 - 4 x + q = 0 . If are in G.P., then the integral values of p and q respectively, are [2001]

Question

Let $α, β$ be the roots of $x^{2} - x + p = 0$ and $γ, δ$ be the roots of $x^{2} - 4 x + q = 0$ . If $α, β, γ, δ$ are in G.P., then the integral values of $p$ and $q$ respectively, are [2001]

Smart Achievers · Accepted Answer

(1)

$α, β are the roots of x^{2} - x + p = 0$

$∴$ $x^{2} - 4 x + q = 0$

$α β = p . . . (i i)$

$γ, δ are the roots of x^{2} - 4 x + q = 0$

$∴$ $γ + δ = 4 . . . (i i i)$

$γ δ = q . . . (i v)$

$α, β, γ, δ are in G.P.$

$∴$ $Let α = a, β = a r, γ = a r^{2}, δ = a r^{3}$

$Substituting these values in equations (i), (ii), (iii) and (iv), we get$

$a + a r = 1 . . . (v)$
$a^{2} r = p . . . (v i)$
$a r^{2} + a r^{3} = 4 . . . (v i i)$
$a^{2} r^{5} = q . . . (v i i i)$

$On dividing (vii) by (v), we get$

$\frac{a r^{2} (1 + r)}{a (1 + r)} = \frac{4}{1} \Rightarrow r^{2} = 4 \Rightarrow r = 2, - 2$

$From (v), a = \frac{1}{1 + r} = \frac{1}{1 + 2} or \frac{1}{1 - 2} = \frac{1}{3} or - 1$

Since $p$ is an integer (given), $r$ is also an integer $(2 or - 2)$

$From (vi), a \neq \frac{1}{3} . Hence a = - 1 and r = - 2$

$∴$ $p = {(- 1)}^{2} \times (- 2) = - 2$

$q = {(- 1)}^{2} \times {(- 2)}^{5} = - 32$

Solution

Q.
Let $α, β$ be the roots of $x^{2} - x + p = 0$ and $γ, δ$ be the roots of $x^{2} - 4 x + q = 0$ . If $α, β, γ, δ$ are in G.P., then the integral values of $p$ and $q$ respectively, are [2001]

1 $- 2, - 32$

2 $- 2, 3$

3 $- 6, 3$

4 $- 6, - 32$

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Solution

Q. Let α,β be the roots of x2-x+p=0 and γ,δ be the roots of x2-4x+q=0. If α,β,γ,δ are in G.P., then the integral values of p and q respectively, are [2001]

1 -2,-32

2 -2,3

3 -6,3