Let a , b , c , p , q be real numbers. Suppose are the roots of the equation x 2 + 2 p x + q = 0 and , 1 are the roots of the equation a x 2 + 2 b x + c = 0 , where 2 { - 1 , 0 , 1 } . STATEMENT-1: ( p 2 - q ) ( b 2 - a c ) 0 and STATEMENT-2

Question

Let $a, b, c, p, q$ be real numbers. Suppose $α, β$ are the roots of the equation $x^{2} + 2 p x + q = 0$ and $α, \frac{1}{β}$ are the roots of the equation $a x^{2} + 2 b x + c = 0,$ where $β^{2} \notin {- 1, 0, 1} .$

$STATEMENT-1: (p^{2} - q) (b^{2} - a c) \geq 0$

$and$

$STATEMENT-2: b \neq p a or c \neq q a$ [2008]

$STATEMENT-1: (p^{2} - q) (b^{2} - a c) \geq 0$

$and$

$STATEMENT-2: b \neq p a or c \neq q a$ [2008]

Smart Achievers · Accepted Answer

(2)

As $a, b, c, p, q \in R$ and the two given equations have exactly one common root

$\Rightarrow$ Either both equations have real roots or both equations have imaginary roots

$\Rightarrow Either D_{1} \geq 0 and D_{2} \geq 0 or D_{1} \leq 0 and D_{2} \leq 0$

$\Rightarrow p^{2} - q \geq 0 and b^{2} - a c \geq 0$ $or p^{2} - q \leq 0 and b^{2} - a c \leq 0$

$\Rightarrow (p^{2} - q) (b^{2} - a c) \geq 0$

$∴$ $Statement 1 is true.$

$Also we have α β = q and \frac{α}{β} = \frac{c}{a}$

$∴$ $\frac{α β}{α / β} = \frac{q}{c} \times a \Rightarrow β^{2} = \frac{q a}{c}$

$As β \neq 1 or - 1 \Rightarrow β^{2} \neq 1 \Rightarrow \frac{q a}{c} \neq 1 or c \neq q a$

$Again, as exactly one root α is common, and β \neq 1$

$∴$ $α + β \neq α + \frac{1}{β} \Rightarrow \frac{- 2 b}{a} \neq - 2 p \Rightarrow b \neq a p$

$∴$ $Statement 2 is correct.$

$But Statement 2 is not a correct explanation of Statement 1.$

Solution

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

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

3 STATEMENT - 1 is True, STATEMENT - 2 is False

4 STATEMENT - 1 is False, STATEMENT - 2 is True

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