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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 −2x−3)(3^x −27)=0  does the equation have two roots  or three roots?

$$\mathrm{discussion}\:\mathrm{back}\:\mathrm{with}\:\mathrm{mr}\:\mathrm{W}. \\ $$$$\mathrm{consider}\:\mathrm{this}\:\mathrm{equation}\: \\ $$$$\left(\mathrm{x}^{\mathrm{2}} −\mathrm{2x}−\mathrm{3}\right)\left(\mathrm{3}^{\mathrm{x}} −\mathrm{27}\right)=\mathrm{0} \\ $$$$\mathrm{does}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{have}\:\mathrm{two}\:\mathrm{roots} \\ $$$$\mathrm{or}\:\mathrm{three}\:\mathrm{roots}? \\ $$

Commented by naka3546 last updated on 27/Jan/20

x ∈ R  or  x ∈ C ?

$${x}\:\in\:\mathbb{R}\:\:{or}\:\:{x}\:\in\:\mathbb{C}\:? \\ $$

Commented by john santu last updated on 27/Jan/20

x∈R sir

$$\mathrm{x}\in\mathbb{R}\:\mathrm{sir} \\ $$

Answered by MJS last updated on 27/Jan/20

x^2 −2x−3=0∨3^x −27=0  (x=−1∨x=3)∨(x=3)≡(x=−1∨x=3)  2 roots  we have different kinds of roots, in this case  1 single root x=−1  1 double root x=3

$${x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{3}=\mathrm{0}\vee\mathrm{3}^{{x}} −\mathrm{27}=\mathrm{0} \\ $$$$\left({x}=−\mathrm{1}\vee{x}=\mathrm{3}\right)\vee\left({x}=\mathrm{3}\right)\equiv\left({x}=−\mathrm{1}\vee{x}=\mathrm{3}\right) \\ $$$$\mathrm{2}\:\mathrm{roots} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{different}\:\mathrm{kinds}\:\mathrm{of}\:\mathrm{roots},\:\mathrm{in}\:\mathrm{this}\:\mathrm{case} \\ $$$$\mathrm{1}\:\mathrm{single}\:\mathrm{root}\:{x}=−\mathrm{1} \\ $$$$\mathrm{1}\:\mathrm{double}\:\mathrm{root}\:{x}=\mathrm{3} \\ $$

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

$${f}'\left({x}\right)=\mathrm{2}\left({x}−\mathrm{1}\right)\left(\mathrm{3}^{{x}} −\mathrm{27}\right)+\left({x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{3}\right)\mathrm{3}^{{x}} \mathrm{ln}\:\mathrm{3} \\ $$$${f}'\left(−\mathrm{1}\right)=\frac{\mathrm{320}}{\mathrm{3}}>\mathrm{0}\:\Rightarrow\:\begin{cases}{{f}\left({x}\right)<\mathrm{0};\:{x}<−\mathrm{1}}\\{{f}\left({x}\right)>\mathrm{0};\:{x}>−\mathrm{1}}\end{cases} \\ $$$${f}'\left(\mathrm{3}\right)=\mathrm{0}\:\Rightarrow\:{f}\left({x}\right)>\mathrm{0};\:−\mathrm{1}<{x}<\infty \\ $$$$\mathrm{at}\:{x}=\mathrm{3}\:\mathrm{there}'\mathrm{s}\:\mathrm{also}\:\mathrm{a}\:\mathrm{local}\:\mathrm{minimum} \\ $$$$ \\ $$$$\mathrm{counting}\:\mathrm{roots}\:\mathrm{means}\:\mathrm{to}\:\mathrm{count}\:\mathrm{the}\:{x}−\mathrm{values} \\ $$$$\mathrm{at}\:\mathrm{which}\:{y}\:\mathrm{becomes}\:\mathrm{0} \\ $$$$\mathrm{after}\:\mathrm{counting}\:\mathrm{we}\:\mathrm{classify}\:\mathrm{these}\:\mathrm{roots} \\ $$$${y}=\left({x}−\mathrm{3}\right)^{\mathrm{17}} \:\mathrm{has}\:\mathrm{only}\:\mathrm{one}\:\mathrm{root}\:\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{a} \\ $$$$\mathrm{seventeen}−\mathrm{fold}\:\mathrm{root} \\ $$

Commented by john santu last updated on 27/Jan/20

ok. thank you mister

$$\mathrm{ok}.\:\mathrm{thank}\:\mathrm{you}\:\mathrm{mister} \\ $$

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