The roots of the equation $(3 - x)^{4} + (2 - x)^{4} = (5 - 2 x)^{4}$ are

$2$ real and $2$ non real complex

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All are non real complex

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$3$ are real & one non real complex

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All real

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Solution

The correct option is A
$2$ real and $2$ non real complex

$(3 - x)^{4} + (2 - x)^{4} = (5 - 2 x)^{4}$

Put $a = 3 - x, b = 2 - x$

$\Rightarrow a^{4} + b^{4} = (a + b)^{4}$

$\Rightarrow a^{4} + b^{4} = a^{4} + 4 a^{3} b + 6 a^{2} b^{2} + 4 a b^{3} + b^{4}$

$\Rightarrow 2 a b (2 a^{2} + 3 a b + 2 b^{2}) = 0$

$a = 0$ or $b = 0$ or $2 a^{2} + 3 a b + 2 b^{2} = 0$

i.e., $3 - x = 0$ or $2 - x = 0$ or $7 x^{2} - 35 x + 44 = 0$

Solving we get $x = 3, x = 2, x = \frac{(35 \pm i \sqrt{7})}{14}$

Suggest Corrections

Similar questions

The roots of the equation $(3 - x)^{4} + (2 - x)^{4} = (5 - 2 x)^{4}$ are

x^{4} - 3 \sqrt{2} x^{3} + 3 x^{2} + 3 \sqrt{2 x} + 4,

\sqrt{2}, a n d, 2 \sqrt{2}

5

x^{4} + 1 = 0

(3 - x)^{4} + (5 - x)^{4} = 16

x^{4} - x^{2} + 2 x - 1 = 0

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