The number of real roots of the equation ${(x + 3)}^{4} + {(x + 5)}^{4} = 16$ , is:

0

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2

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4

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none

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Solution

The correct option is B $2$
Given, ${(x + 3)}^{4} + {(x + 5)}^{4} = 16$
$\Rightarrow {(x + 4 - 1)}^{4} + {(x + 4 + 1)}^{4} = 16$
Substituting $y = x + 4$
${(y - 1)}^{4} + {(y + 1)}^{4} = 16 \Rightarrow 2 (y^{4} + 6 y^{2} + 1) = 16 \Rightarrow y^{4} + 6 y^{2} - 7 = 0 \Rightarrow (y^{2} + 7) (y^{2} - 1) = 0 \Rightarrow y = \pm i \sqrt{7}, \pm 1$
For $y = \pm 1 \Rightarrow x + 4 = \pm 1 \Rightarrow x = 3, - 5$
Hence, number of real roots are $2$ .

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