Math Question and Answers


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 24565 by ajfour last updated on 21/Nov/17

$y = {a x}^{3} + {b x}^{2} + c x + d, t h e n$ $p r o v e t h a t t h e e q u a t i o n y = 0$ $h a s o n l y o n e r e a l r o o t i f$ $a [{(9 a d - b c)}^{2} - 4 (b^{2} - 3 a c) (c^{2} - 3 b d)]$ $> 0 p r o v i d e d b^{2} > 3 a c .$
	Answered by ajfour last updated on 21/Nov/17

	Commented byajfour last updated on 21/Nov/17

	$I f t h e l o c a l m i n i m u m v a l u e$ $a n d t h e l o c a l m a x i m u m v a l u e,$ $b o t h, a r e o f t h e s a m e s i g n, t h e n,$ $i b e l i e v e, t h e r e c a n b e j u s t o n e$ $r e a l r o o t o f a c u b i c e q u a t i o n .$ $y = {a x}^{3} + {b x}^{2} + c x + d$ $\Rightarrow \frac{d y}{d x} = 3 {a x}^{2} + 2 b x + c$ $l e t a t x = α, β \frac{d y}{d x} = 0$ $\Rightarrow α β = \frac{c}{3 a} a n d (α + β) = - \frac{2 b}{3 a}$ $\Rightarrow 3 a α^{2} + 2 b α + c = 0 . . . . . (i)$ $3 a β^{2} + 2 b β + c = 0 . . . . . (i i)$ $F o r o n e r e a l r o o t$ $y (α) \times y (β) > 0$ $o r 3 y (α) \times 3 y (β) > 0$ $3 y (α) = 3 a α^{3} + 3 b α^{2} + 3 c α + 3 d$ $s u b t r a c t i n g α \times (i) f r o m t h i s$ $3 y (α) = b α^{2} + 2 c α + 3 d$ $= \frac{b}{3 a} (3 a α^{2}) + 2 c α + 3 d$ $u s i n g (i) a g a i n :$ $3 y (α) = - \frac{b}{3 a} (2 b α + c) + 2 c α + 3 d$ $= 2 α (c - \frac{b^{2}}{3 a}) + (3 d - \frac{b c}{3 a})$ $s o 3 y (α) \times 3 y (β) =$ $[4 α β {(c - \frac{b^{2}}{3 a})}^{2} + 2 (α + β) (c - \frac{b^{2}}{3 a}) (3 d - \frac{b c}{3 a})$ $+ {(3 d - \frac{b c}{3 a})}^{2}]$ $A s α = \frac{c}{3 a} a n d β = - \frac{2 b}{3 a} w e h a v e$ $3 y (α) \times 3 y (β) =$ $\frac{4 c}{3 a} {(c - \frac{b^{2}}{3 a})}^{2} - \frac{4 b}{3 a} (c - \frac{b^{2}}{3 a}) (3 d - \frac{b c}{3 a})$ $+ {(3 d - \frac{b c}{3 a})}^{2} > 0$ $o r$ $4 c {(3 a c - b^{2})}^{2} - 4 b (3 a c - b^{2}) (9 a d - b c)$ $+ 3 a {(9 a d - b c)}^{2} > 0$ $o r$ $3 a {(9 a d - b c)}^{2} + 4 (3 a c - b^{2}) (3 {a c}^{2} - b^{2} c$ $- 9 a b d + b^{2} c) > 0$ $\Rightarrow$ $3 a {(9 a d - b c)}^{2} + 4 (3 a) (3 a c - b^{2}) (c^{2} - 3 b d) > 0$ $a [{(9 a d - b c)}^{2} - 4 (b^{2} - 3 a c) (c^{2} - 3 b d)] > 0 .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum