Let a,b,c,p,q be real numbers, suppose α,β are the roots of equation x2+2px+q=0 and α,1β are roots of equation ax2+2bx+c=0 where β2∉{−1,0,1} Statement I: (p2−q)(b2−ac)≥0Statement II: b≠pa or c≠qa

(2) If the roots are imaginary, then β=α¯,1β=α¯⇒β2=1 , contradiction ∴ The roots are real ⇒(p2−q) and (b2−ac)≥0suppose b = pa and c= qa then the second equation becomes identical with the first equation.∴β=1β⇒β2=1 , contradiction∴ Either b≠pa or c≠qa .

Let a,b,c,p,q be real numbers, suppose α,β are the roots of equation x2+2px+q=0 and α,1β are roots of equation ax2+2bx+c=0 where β2∉{−1,0,1} Statement I: (p2−q)(b2−ac)≥0Statement II: b≠pa or c≠qa

(2) If the roots are imaginary, then β=α¯,1β=α¯⇒β2=1 , contradiction ∴ The roots are real ⇒(p2−q) and (b2−ac)≥0suppose b = pa and c= qa then the second equation becomes identical with the first equation.∴β=1β⇒β2=1 , contradiction∴ Either b≠pa or c≠qa .

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Let $a, b, c, p, q$ be real numbers, suppose $α, β$ are the roots of equation $x^{2} + 2 p x + q = 0$ and $α, \frac{1}{β}$ are roots of equation $a x^{2} + 2 b x + c = 0$ where $β^{2} \notin {- 1, 0, 1}$
Statement I: $(p^{2} - q) (b^{2} - a c) \geq 0$
Statement II: $b \neq p a o r c \neq q a$

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Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

Statement-1 is True, Statement-2 is False

Statement-1 is False, Statement-2 is True

answer is B.

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Detailed Solution

(2) If the roots are imaginary, then    $β = \bar{α}, \frac{1}{β} = \bar{α} \Rightarrow β^{2} = 1$ , contradiction
$∴$ The roots are real
   $\Rightarrow (p^{2} - q)$ and $(b^{2} - a c) \geq 0$
suppose b = pa and c= qa then the second equation becomes identical with the first equation.
$∴ β = \frac{1}{β} \Rightarrow β^{2} = 1$ , contradiction
$∴$ Either    $b \neq p a$ or $c \neq q a$ .