The roots of the equation (3−x)^4 +(2−x)^4 =(5−2x)^4 are (a) all real (b) all imaginary (c) two real and two imaginary (d)none of the above .


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Question Number 20552 by ajfour last updated on 28/Aug/17

$T h e r o o t s o f t h e e q u a t i o n$ ${(3 - x)}^{4} + {(2 - x)}^{4} = {(5 - 2 x)}^{4} a r e$ $(a) a l l r e a l (b) a l l i m a g i n a r y$ $(c) t w o r e a l a n d t w o i m a g i n a r y$ $(d) n o n e o f t h e a b o v e .$
	Answered by Tinkutara last updated on 28/Aug/17

	${(x - 3)}^{4} + {(x - 2)}^{4} = {(2 x - 5)}^{4}$ $L e t a = x - 3, b = x - 2, t h e n s o l v i n g f o r$ $a^{4} + b^{4} = {(a + b)}^{4}$ $a b (2 a^{2} + 2 b^{2} + 3 a b) = 0$ $O n e s o l u t i o n i s a b = 0 \Rightarrow (x - 3) (x - 2) = 0$ $\Rightarrow x = 2, 3 (T w o i n t e g r a l r o o t s)$ $a n d 2 {(a + b)}^{2} = a b$ $\Rightarrow 2 {(2 x - 5)}^{2} = x^{2} - 5 x + 6$ $2 (4 x^{2} - 20 x + 25) = x^{2} - 5 x + 6$ $7 x^{2} - 35 x + 44 = 0$ $D = 1225 - 4 \times 7 \times 44 = - 7 < 0$ $H e n c e 2 r e a l a n d 2 i m a g i n a r y r o o t s .$
	Commented by ajfour last updated on 28/Aug/17

	$E x c e l l e n t! k e e p i t u p .$
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