The number of real roots of the equation $(x + 3)^{2} + (x + 1)^{2} + (x - 5)^{2} + (x - 6)^{2} = 0$ is

0

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2

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1

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N o n e o f t h e s e

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Solution

The correct option is A $0$
Given that,
$(x + 3)^{2} + (x + 1)^{2} + (x - 5)^{2} + (x - 6)^{2} = 0$

$⟹ x^{2} + 6 x + 9 + x^{2} + 2 x + 1 + x^{2} - 10 x + 25 + x^{2} - 12 x + 36 = 0$

$⟹ 4 x^{2} - 14 x + 71 = 0$

$∴ b^{2} - 4 a c = (- 14)^{2} - 4 (4) (71) = 196 - 1136 = - 940 < 0$

Hence, given equation has no real roots

Option A is correct.

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