Find the value of m given that the roots of x^4 −15x^3 +70x^2 −120x+m=0 form a geometric progression.


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Question Number 198932 by necx122 last updated on 25/Oct/23

$F i n d t h e v a l u e o f m g i v e n t h a t t h e$ $r o o t s o f x^{4} - 15 x^{3} + 70 x^{2} - 120 x + m = 0$ $f o r m a g e o m e t r i c p r o g r e s s i o n .$
	Answered by deleteduser1 last updated on 25/Oct/23

	$m = a b c d = ({b r}^{- 1}) b (b r) ({b r}^{2}) = b^{4} r^{2}$ $a + b + c + d = 15 \Rightarrow a (1 + r + r^{2} + r^{3}) = 15 . . . (i)$ $a b + b c + c d + d a + c a + b d = 70$ $\Rightarrow a^{2} (r + r^{3} + r^{5} + r^{3} + r^{2} + r^{4}) = 70 . . . (i i)$ $a b c + b c d + a c d + a b d = b^{3} (1 + r^{3} + r^{2} + r) = 120 . . . (i i i)$ $(i) & (i i i) \Rightarrow \frac{120}{b^{3}} = \frac{15}{a} = 8 a = b^{3}$ $b^{3} = a^{3} r^{3} = 8 a \Rightarrow a^{2} r^{3} = 8$ $\Rightarrow m = b^{4} r^{2} = a^{4} r^{6} = {(a^{2} r^{3})}^{2} = 64$
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