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Discriminant

Let α and ...

Question

Let $α$ and $β$ be nonzero real roots of the quadratic equation $x^{2} + a x + b = 0$ and $α + β, α - β, - α + β$ and $- α - β$ be the roots of the equation $x^{4} + a x^{3} + c x^{2} + d x + e = 0$ Then which of the following statement is false?

A

a = 0

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B

c = 0

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C

d = 0

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D

e = 0

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Solution

The correct option is B $c = 0$
Given: $α, β$ are roots of the equation $x^{2} + a x + b = 0$ and $(α + β), (α - β), (β - α), (- (α + β))$ are roots of the equation $x^{4} + a x^{3} + c x^{2} + d x + e = 0$
To find which of the given statement is false
Sol: As $α, β$ are roots of the equation $x^{2} + a x + b = 0$
Hence $α + β = - a, α β = b . . . . . . . . . . (i)$
$(α + β), (α - β), (β - α), (- (α + β))$ are roots of the equation $x^{4} + a x^{3} + c x^{2} + d x + e = 0$
Hence, $(α + β) + (α - β) + (β - α) + (- (α + β)) = - a ⟹ a = 0... (i i)$
And,
$(α + β) (α - β) + (α + β) (β - α) + (α + β) (- (α + β)) + (α - β) (β - α) + (α - β) (- (α + β)) + (β - α) (- (α + β)) = c$
$⟹ (0) (α - β) + (0) (β - α) + (0) (- (0)) + (α - β) (β - α) + (α - β) (- (0)) + (β - α) (- (0)) = c$
[as $α + β = - a = 0$ from (i) and (ii)]
$⟹ (α - β) (β - α) = c$
$⟹ α β - α^{2} - β^{2} - α β = c ⟹ - (α^{2} + β^{2}) = c$
$⟹ - [(α + β)^{2} - 2 α β] = c$ [as $a^{2} + b^{2} = (a + b)^{2} - 2 a b$ ]
$⟹ - (0^{2} - 2 (b)) = c ⟹ c = 2 b$
And,
$(α + β) (α - β) (β - α) + (α + β) (α - β) (- (α + β)) + (α + β) (β - α) (- (α + β)) + (α - β) (β - α) (- (α + β)) = - d ⟹ (0) (α - β) (β - α) + (0) (α - β) (- (0)) + (0) (β - α) (- (0)) + (α - β) (β - α) (- (0)) = - d$
[as $α + β = - a = 0$ from (i) and (ii)]
$⟹ d = 0$
And
$(α + β) (α - β) (β - α) (- (α + β)) = e ⟹ e = 0$
[as $α + β = - a = 0$ from (i) and (ii)]

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