discussion back with mr W. consider this equation (x^2 −2x−3)(3^x −27)=0 does the equation have two roots or three roots?


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Question Number 79647 by john santu last updated on 27/Jan/20

$discussion back with mr W .$ $consider this equation$ $(x^{2} - 2 x - 3) (3^{x} - 27) = 0$ $does the equation have two roots$ $or three roots ?$
Commented by naka3546 last updated on 27/Jan/20

$x \in R o r x \in C ?$
Commented by john santu last updated on 27/Jan/20

$x \in R sir$
	Answered by MJS last updated on 27/Jan/20

	$x^{2} - 2 x - 3 = 0 \lor 3^{x} - 27 = 0$ $(x = - 1 \lor x = 3) \lor (x = 3) \equiv (x = - 1 \lor x = 3)$ $2 roots$ $we have different kinds of roots, in this case$ $1 single root x = - 1$ $1 double root x = 3$
	Commented by MJS last updated on 27/Jan/20
	$f′(x)=2(x−1)(3^x −27)+(x^2 −2x−3)3^x ln 3 f′(−1)=((320)/3)>0 ⇒ { ((f(x)<0; x<−1)),((f(x)>0; x>−1)) :} f′(3)=0 ⇒ f(x)>0; −1<x<∞ at x=3 there′s also a local minimum counting roots means to count the x−values at which y becomes 0 after counting we classify these roots y=(x−3)^(17) has only one root but it′s a seventeen−fold root$
	$f^{'} (x) = 2 (x - 1) (3^{x} - 27) + (x^{2} - 2 x - 3) 3^{x} \ln 3$ $f^{'} (- 1) = \frac{320}{3} > 0 \Rightarrow {\begin{cases} f (x) < 0; x < - 1 \\ f (x) > 0; x > - 1 \end{cases}$ $f^{'} (3) = 0 \Rightarrow f (x) > 0; - 1 < x < \infty$ $at x = 3 {there}^{'} s also a local minimum$ $counting roots means to count the x - values$ $at which y becomes 0$ $after counting we classify these roots$ $y = {(x - 3)}^{17} has only one root but {it}^{'} s a$ $seventeen - fold root$
	Commented by john santu last updated on 27/Jan/20

	$ok . thank you mister$
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