Q#141663 by ajfour sir reposted. x^2 (x−12)(x−15)=k(x−16) ;k>0 Find x in terms of k.


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Question Number 142939 by Rasheed.Sindhi last updated on 07/Jun/21

$You can't use 'macro parameter character #' in math mode$ $x^{2} (x - 12) (x - 15) = k (x - 16); k > 0$ $F i n d x i n t e r m s o f k .$
	Answered by Rasheed.Sindhi last updated on 07/Jun/21
	${x^2 (x−12)(x−15)}/(x−16)=k Quotient is k with no remainder. Let p(x)=x^2 (x^2 −27x+180) =x^4 −27x^3 +180x^2 +0x+0 By synthetic division: determinant (((16)),1,(−27),(180),0,0),(,,( 16),(−176),(64),(1024)),(,1,(−11),( 4),(64),(1024))) p(x)=Q(x)D(x)+R x^2 (x−12)(x−15)= (x^3 −11x^2 +4x+64)(x−16)+1024 Neither the quotient is constant nor the remainder is zero ∀ x So this proves that the equation has no solution.$
	${x^{2} (x - 12) (x - 15)} / (x - 16) = k$ $Q u o t i e n t i s k w i t h n o r e m a i n d e r .$ $L e t p (x) = x^{2} (x^{2} - 27 x + 180)$ $= x^{4} - 27 x^{3} + 180 x^{2} + 0 x + 0$ $B y s y n t h e t i c d i v i s i o n :$ $\begin{array}{ccc} 16) & 1 & - 27 & 180 & 0 & 0 \\ 16 & - 176 & 64 & 1024 \\ 1 & - 11 & 4 & 64 & 1024 \end{array}$ $p (x) = Q (x) D (x) + R$ $x^{2} (x - 12) (x - 15) =$ $(x^{3} - 11 x^{2} + 4 x + 64) (x - 16) + 1024$ $N e i t h e r t h e q u o t i e n t i s c o n s t a n t$ $n o r t h e r e m a i n d e r i s z e r o \forall x$ $S o t h i s p r o v e s t h a t t h e e q u a t i o n$ $h a s n o s o l u t i o n .$
	Commented bymr W last updated on 07/Jun/21

	$i t h i n k {i t}^{'} s w r o n g s o m e w h e r e s i r .$ $y = x^{2} (x - 12) (x - 15) i s a c u r v e p a s s i n g$ $t h r o u g h (0, 0), (12, 0) a n d (15, 0)$ $y = k (x - 16) i s a l i n e p a s s i n g t h r o u g h$ $(16, 0) w i t h i n c l i n a t i o n k .$ $f r o m (16, 0) w e c a n d r a w t w o t a n g e n t$ $l i n e s t o t h e c u r v e, s a y t h e s e t a n g e n t$ $l i n e s h a v e i n c l i n a t i o n k_{1} a n d k_{2} .$ $w e k n o w f o r k ⩽ k_{1} o r k ⩾ k_{2}, t h e l i n e$ $y = k (x - 16) i n t e r s e c t s t h e c u r v e a t$ $l e a s t a t o n e p o i n t a n d a t m o s t a t$ $f o u r p o i n t s . t h a t m e a n s f o r$ $k ⩽ k_{1} o r k ⩾ k_{2}, t h e e q u a t i o n$ $x^{2} (x - 12) (x - 15) = k (x - 16) h a s$ $a t l e a s t o n e s o l u t i o n a n d a t m o s t$ $f o u r s o l u t i o n s .$
	Commented bymr W last updated on 07/Jun/21

	Commented byRasheed.Sindhi last updated on 08/Jun/21

	$T h a n X f o r y o u r E f f o r t s S i r!$ $P e r h a p s m y l o g i c i s o n l y v a l i d f o r$ $i n t e g e r k (?) .$
	Commented bymr W last updated on 08/Jun/21

	$t h e q u e s t i o n {d o e s n}^{'} t r e q u e s t i f$ $x^{2} (x - 12) (x - 15) c a n b e e x p r e s s e d a s$ $k (x - 16) f o r a n y v a l u e s o f x, b u t o n l y$ $a s k s i f o n e o r m o r e v a l u e s o f x e x i s t$ $s u c h t h a t x^{2} (x - 12) (x - 15) = k (x - 16) .$
	Commented byRasheed.Sindhi last updated on 08/Jun/21

	$OK S i r, T h a n X a g a i n .$
	Answered by Rasheed.Sindhi last updated on 07/Jun/21

	$x^{2} (x - 12) (x - 15) = k (x - 16); k > 0$ $T h i s m e a n s :$ $W h e n x^{2} (x - 12) (x - 15) i s d i v i d e d$ $b y x - 16, T h e q u o t i e n t i s k (> 0) a n d$ $t h e r e m a i n d e r i s 0$ $B u t w e c a n v e r i f y t h a t t h e r e m a i n d e r$ $R i s :$ $R = {(16)}^{2} (16 - 12) (16 - 15)$ $= 256.4.1 = 1024 \neq 0$ $S o f o r a n y v a l u e o f x t h e r e m a i n d e r$ ${c a n}^{'} t b e z e r o .$ $T h i s p r o v e s t h a t t h e a b o v e e q u a t i o n$ ${c a n}^{'} t b e s o l v e d (e v e n i n t e r m s o f k > 0 .)$
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