A quadratic equation p(x)=0 having coefficient of x^2 unity is such that p(x)=0 and p(p(p(x)))=0 have a common root then, prove that : p(0)×p(1)=0.


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 31804 by rahul 19 last updated on 15/Mar/18

$A q u a d r a t i c e q u a t i o n p (x) = 0 h a v i n g$ $c o e f f i c i e n t o f x^{2} u n i t y i s s u c h t h a t$ $p (x) = 0 a n d p (p (p (x))) = 0 h a v e a$ $c o m m o n r o o t t h e n,$ $p r o v e t h a t : p (0) \times p (1) = 0 .$
	Answered by MJS last updated on 15/Mar/18

	$p (x) = x^{2} + b x + c$ $p (0) \times p (1) = 0 \Rightarrow p (0) = 0 \lor p (1) = 0$ $case 1$ $p (0) = 0 \Rightarrow c = 0$ $p (x) = (x + b) x$ $p (p (x)) = q (x)$ $p (p (p (x))) = p (q (x)) = r (x)$ $q (x) = (p (x) + b) p (x) \Rightarrow$ $\Rightarrow q (0) = 0$ $r (x) = (q (x) + b) q (x) \Rightarrow$ $\Rightarrow r (0) = 0 \Rightarrow$ $\Rightarrow 0 = common root of p (x) and r (x)$ $case 2$ $p (1) = 0$ $p (x) = (x - 1) (x - c)$ $[with b = - c - 1; p (x) = x^{2} - (c + 1) x + c]$ $q (x) = (p (x) - 1) (p (x) - c) =$ $= p^{2} (x) - (c + 1) p (x) + c$ $r (x) = [p^{2} (x) - (c + 1) p (x) + c - 1] \times$ $\times [p^{2} (x) - (c + 1) p (x)] =$ $= [. . .] \times [p (x) - (c + 1)] \times p (x) \Rightarrow$ $\Rightarrow r (1) = 0 \Rightarrow$ $\Rightarrow 1 = common root of p (x) and r (x)$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum