First slide

Theory of equations

Question

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

Moderate

A

all real
B

all imaginary
C

two real and two imaginary
D

none of the above

Solution

The given equation is ${(3 - x)}^{4} + {(2 - x)}^{4} = {(5 - 2 x)}^{4}$         ……(i)
Put $3 - x = a$ and $2 - x = b$

Eq. (i) be comes $a^{4} + b^{4} = {(a + b)}^{4}$

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

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

$∴$ $a = 0 \Rightarrow 3 - x = 0 \Rightarrow x = 3$

      $b = 0 \Rightarrow 2 - x = 0 \Rightarrow x = 2$

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

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

       $\frac{a}{b} = \frac{- 3 \pm \sqrt{9 - 16}}{4}$

      $\frac{3 - x}{2 - x} = \frac{- 3 \pm i \sqrt{7}}{4}$

$\Rightarrow$ $\frac{3 - x}{2 - x} - 1 = \frac{- 7 \pm i \sqrt{7}}{4}$

$\Rightarrow$ $\frac{1}{2 - x} = \frac{- 7 \pm i \sqrt{7}}{4}$

    $2 - x = \frac{4}{- 7 \pm i \sqrt{7}}$

    $x - 2 = \frac{4}{- 7 \pm i \sqrt{7}}$

       $x = \frac{- 18 \pm 2 i \sqrt{7}}{- 7 \pm i \sqrt{7}}$

Taking ‘+’ sign

       $x = \frac{(- 18 + 2 i \sqrt{7}) (- 7 - i \sqrt{7})}{(- 7 + i \sqrt{7}) (- 7 - i \sqrt{7})}$

          $= \frac{140 + 4 i \sqrt{7}}{56}$

$∴$ $x = \frac{35 + i \sqrt{7}}{14}$

Taking ‘-‘ sign,    $x = \frac{35 - i \sqrt{7}}{14}$

Hence, roots of Eq. (i) are $2, 3, \frac{35 \pm i \sqrt{7}}{14}$

i.e.,   Two real and two imaginary.

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