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Q1-Evaluate-1-1-x-x-3-1-dx-Q2-Find-the-sum-of-all-integers-k-for-which-the-equation-2x-3-6x-2-k-0-has-more-than-one-solution-Q3-Find-the-shortest-distance-from-a-point-on-the-c

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Question Number 103492 by abony1303 last updated on 15/Jul/20
Q1: Evaluate ∫_(−1) ^( 1) ∣x∣∙(x^3 +1)dx Q2: Find the sum of all integers k for which the equation 2x^3 −6x^2 +k=0 has more than one solution. Q3: Find the shortest distance from a point on the curve y=x^2 −x to the line y=x−3
$$\mathrm{Q1}:\:\:\mathrm{Evaluate}\:\int_{−\mathrm{1}} ^{\:\mathrm{1}} \mid\mathrm{x}\mid\centerdot\left(\mathrm{x}^{\mathrm{3}} +\mathrm{1}\right)\mathrm{dx} \\ $$$$ \\ $$$$\mathrm{Q2}:\:\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{all}\:\mathrm{integers}\:{k}\:\mathrm{for}\: \\ $$$$\mathrm{which}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{2}{x}^{\mathrm{3}} −\mathrm{6}{x}^{\mathrm{2}} +{k}=\mathrm{0} \\ $$$$\mathrm{has}\:\mathrm{more}\:\mathrm{than}\:\mathrm{one}\:\mathrm{solution}. \\ $$$$ \\ $$$$\mathrm{Q3}:\:\mathrm{Find}\:\mathrm{the}\:\mathrm{shortest}\:\mathrm{distance}\:\mathrm{from}\:\mathrm{a}\: \\ $$$$\mathrm{point}\:\mathrm{on}\:\mathrm{the}\:\mathrm{curve}\:{y}={x}^{\mathrm{2}} −{x}\:\mathrm{to}\:\mathrm{the}\:\mathrm{line} \\ $$$${y}={x}−\mathrm{3} \\ $$
Commented by abony1303 last updated on 15/Jul/20
pls help
$$\mathrm{pls}\:\mathrm{help} \\ $$
Answered by mr W last updated on 15/Jul/20
Q2 2x^2 (x−3)+k=0 k<0 ⇒one root: x>3 k=0 ⇒two roots: x=0 or 3 k>0: (1/x^3 )−(6/(kx))+(2/k)=0 such that three roots: ((1/k))^2 +(−(2/k))^3 <0 ((1/k))^2 (1−(8/k))<0 1−(8/k)<0 k<8 ⇒k=1,2,3,...,7 Σk=((7(1+7))/2)=28
$${Q}\mathrm{2} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} \left({x}−\mathrm{3}\right)+{k}=\mathrm{0} \\ $$$${k}<\mathrm{0}\:\Rightarrow{one}\:{root}:\:{x}>\mathrm{3} \\ $$$${k}=\mathrm{0}\:\Rightarrow{two}\:{roots}:\:{x}=\mathrm{0}\:{or}\:\mathrm{3} \\ $$$${k}>\mathrm{0}: \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{3}} }−\frac{\mathrm{6}}{{kx}}+\frac{\mathrm{2}}{{k}}=\mathrm{0} \\ $$$${such}\:{that}\:{three}\:{roots}: \\ $$$$\left(\frac{\mathrm{1}}{{k}}\right)^{\mathrm{2}} +\left(−\frac{\mathrm{2}}{{k}}\right)^{\mathrm{3}} <\mathrm{0} \\ $$$$\left(\frac{\mathrm{1}}{{k}}\right)^{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{8}}{{k}}\right)<\mathrm{0} \\ $$$$\mathrm{1}−\frac{\mathrm{8}}{{k}}<\mathrm{0} \\ $$$${k}<\mathrm{8} \\ $$$$\Rightarrow{k}=\mathrm{1},\mathrm{2},\mathrm{3},…,\mathrm{7} \\ $$$$\Sigma{k}=\frac{\mathrm{7}\left(\mathrm{1}+\mathrm{7}\right)}{\mathrm{2}}=\mathrm{28} \\ $$
Commented by abony1303 last updated on 15/Jul/20
thank you sir. but I drew a graph and k=8 also correct? And pls can you explain the raw such that three roots... and below it.
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{but}\:\mathrm{I}\:\mathrm{drew}\:\mathrm{a}\:\mathrm{graph}\:\mathrm{and}\:\mathrm{k}=\mathrm{8} \\ $$$$\mathrm{also}\:\mathrm{correct}?\:\mathrm{And}\:\mathrm{pls}\:\mathrm{can}\:\mathrm{you}\:\mathrm{explain}\:\mathrm{the} \\ $$$$\mathrm{raw}\:{such}\:{that}\:{three}\:{roots}…\:{and}\:\mathrm{below}\:{it}. \\ $$
Commented by mr W last updated on 15/Jul/20
i misread the question as: more than two roots. for more than one root: the answer is Σk=0+1+2+3+...+8=36 we have: (1/x^3 )−(6/(kx))+(2/k)=0 let t=(1/x), then t^3 −(6/k)t+(2/k)=0 three roots for x ⇒ three roots for t. let f(t)=t^3 +bt+c if f(t)=0 has three roots, it means f′(t)=0 ⇒3t^2 +b=0⇒t_(1,2) =±(√(−(b/3))) f(t_1 )f(t_2 )<0, see diagram. f(t_1 )=t_1 ^3 +bt_1 +c=(1/3)t_1 (3t_1 ^2 +b)+((2bt_1 )/3)+c=((2bt_1 )/3)+c f(t_2 )=((2bt_2 )/3)+c (((2bt_1 )/3)+c)(((2bt_2 )/3)+c)<0 ((4b^2 )/9)t_1 t_2 +((2b)/3)(t_1 +t_2 )+c^2 <0 ((4b^2 )/9)×(b/3)+((2b)/3)×0+c^2 <0 ⇒((b/3))^3 +((c/2))^2 <0 with b=−(6/k), c=(2/k) ⇒(−(2/k))^3 +((1/k))^2 <0
$${i}\:{misread}\:{the}\:{question}\:{as}:\:{more}\:{than} \\ $$$${two}\:{roots}. \\ $$$${for}\:{more}\:{than}\:{one}\:{root}:\:{the}\:{answer} \\ $$$${is}\:\Sigma{k}=\mathrm{0}+\mathrm{1}+\mathrm{2}+\mathrm{3}+…+\mathrm{8}=\mathrm{36} \\ $$$$ \\ $$$${we}\:{have}: \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{3}} }−\frac{\mathrm{6}}{{kx}}+\frac{\mathrm{2}}{{k}}=\mathrm{0} \\ $$$${let}\:{t}=\frac{\mathrm{1}}{{x}},\:{then} \\ $$$${t}^{\mathrm{3}} −\frac{\mathrm{6}}{{k}}{t}+\frac{\mathrm{2}}{{k}}=\mathrm{0} \\ $$$${three}\:{roots}\:{for}\:{x}\:\Rightarrow\:{three}\:{roots}\:{for}\:{t}. \\ $$$$ \\ $$$${let}\:{f}\left({t}\right)={t}^{\mathrm{3}} +{bt}+{c} \\ $$$${if}\:{f}\left({t}\right)=\mathrm{0}\:{has}\:{three}\:{roots},\:{it}\:{means} \\ $$$${f}'\left({t}\right)=\mathrm{0}\:\Rightarrow\mathrm{3}{t}^{\mathrm{2}} +{b}=\mathrm{0}\Rightarrow{t}_{\mathrm{1},\mathrm{2}} =\pm\sqrt{−\frac{{b}}{\mathrm{3}}} \\ $$$${f}\left({t}_{\mathrm{1}} \right){f}\left({t}_{\mathrm{2}} \right)<\mathrm{0},\:{see}\:{diagram}. \\ $$$${f}\left({t}_{\mathrm{1}} \right)={t}_{\mathrm{1}} ^{\mathrm{3}} +{bt}_{\mathrm{1}} +{c}=\frac{\mathrm{1}}{\mathrm{3}}{t}_{\mathrm{1}} \left(\mathrm{3}{t}_{\mathrm{1}} ^{\mathrm{2}} +{b}\right)+\frac{\mathrm{2}{bt}_{\mathrm{1}} }{\mathrm{3}}+{c}=\frac{\mathrm{2}{bt}_{\mathrm{1}} }{\mathrm{3}}+{c} \\ $$$${f}\left({t}_{\mathrm{2}} \right)=\frac{\mathrm{2}{bt}_{\mathrm{2}} }{\mathrm{3}}+{c} \\ $$$$\left(\frac{\mathrm{2}{bt}_{\mathrm{1}} }{\mathrm{3}}+{c}\right)\left(\frac{\mathrm{2}{bt}_{\mathrm{2}} }{\mathrm{3}}+{c}\right)<\mathrm{0} \\ $$$$\frac{\mathrm{4}{b}^{\mathrm{2}} }{\mathrm{9}}{t}_{\mathrm{1}} {t}_{\mathrm{2}} +\frac{\mathrm{2}{b}}{\mathrm{3}}\left({t}_{\mathrm{1}} +{t}_{\mathrm{2}} \right)+{c}^{\mathrm{2}} <\mathrm{0} \\ $$$$\frac{\mathrm{4}{b}^{\mathrm{2}} }{\mathrm{9}}×\frac{{b}}{\mathrm{3}}+\frac{\mathrm{2}{b}}{\mathrm{3}}×\mathrm{0}+{c}^{\mathrm{2}} <\mathrm{0} \\ $$$$\Rightarrow\left(\frac{{b}}{\mathrm{3}}\right)^{\mathrm{3}} +\left(\frac{{c}}{\mathrm{2}}\right)^{\mathrm{2}} <\mathrm{0} \\ $$$${with}\:{b}=−\frac{\mathrm{6}}{{k}},\:{c}=\frac{\mathrm{2}}{{k}} \\ $$$$\Rightarrow\left(−\frac{\mathrm{2}}{{k}}\right)^{\mathrm{3}} +\left(\frac{\mathrm{1}}{{k}}\right)^{\mathrm{2}} <\mathrm{0} \\ $$
Commented by mr W last updated on 15/Jul/20
Commented by mr W last updated on 15/Jul/20
Commented by mr W last updated on 15/Jul/20
Commented by mr W last updated on 15/Jul/20
Commented by mr W last updated on 15/Jul/20
Commented by mr W last updated on 15/Jul/20
Commented by mr W last updated on 15/Jul/20
three roots only for 1≤k≤7. two roots only for k=0 or 8. one root only for k≤−1 or k≥9.
$${three}\:{roots}\:{only}\:{for}\:\mathrm{1}\leqslant{k}\leqslant\mathrm{7}. \\ $$$${two}\:{roots}\:{only}\:{for}\:{k}=\mathrm{0}\:{or}\:\mathrm{8}. \\ $$$${one}\:{root}\:{only}\:{for}\:{k}\leqslant−\mathrm{1}\:{or}\:{k}\geqslant\mathrm{9}. \\ $$
Answered by mr W last updated on 15/Jul/20
Q3 y=x^2 −x y′=2x−1=1 ⇒x=1, y=0 d_(min) =((∣1×1−1×0−3∣)/( (√(1^2 +(−1)^2 ))))=(√2)
$${Q}\mathrm{3} \\ $$$${y}={x}^{\mathrm{2}} −{x} \\ $$$${y}'=\mathrm{2}{x}−\mathrm{1}=\mathrm{1}\:\Rightarrow{x}=\mathrm{1},\:{y}=\mathrm{0} \\ $$$${d}_{{min}} =\frac{\mid\mathrm{1}×\mathrm{1}−\mathrm{1}×\mathrm{0}−\mathrm{3}\mid}{\:\sqrt{\mathrm{1}^{\mathrm{2}} +\left(−\mathrm{1}\right)^{\mathrm{2}} }}=\sqrt{\mathrm{2}} \\ $$
Answered by mr W last updated on 15/Jul/20
Q1 ∫_(−1) ^( 1) ∣x∣∙(x^3 +1)dx =∫_(−1) ^( 1) ∣x∣∙x^3 dx+∫_(−1) ^( 1) ∣x∣dx =0+2∫_0 ^( 1) xdx =1
$${Q}\mathrm{1} \\ $$$$\int_{−\mathrm{1}} ^{\:\mathrm{1}} \mid\mathrm{x}\mid\centerdot\left(\mathrm{x}^{\mathrm{3}} +\mathrm{1}\right)\mathrm{dx} \\ $$$$=\int_{−\mathrm{1}} ^{\:\mathrm{1}} \mid\mathrm{x}\mid\centerdot\mathrm{x}^{\mathrm{3}} \mathrm{dx}+\int_{−\mathrm{1}} ^{\:\mathrm{1}} \mid\mathrm{x}\mid\mathrm{dx} \\ $$$$=\mathrm{0}+\mathrm{2}\int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{xdx} \\ $$$$=\mathrm{1} \\ $$
Answered by OlafThorendsen last updated on 15/Jul/20
Q1. I = ∫_(−1) ^1 ∣x∣(x^3 +1)dx I = ∫_(−1) ^0 −x(x^3 +1)dx+∫_0 ^1 x(x^3 +1)dx I = [−(x^5 /5)−(x^2 /2)]_(−1) ^0 +[(x^5 /5)+(x^2 /2)]_0 ^1 I = (−(1/5)+(1/2))+((1/5)+(1/2)) I = 1 Q2. Δ′ = (−3)^2 −(2)(k) = 9−2k More than one solution ⇔ Δ′>0 ⇔ k<(9/2) ⇔ k∈{0;1;2;3;4} ⇒ sum = 0+1+2+3+4 = 10 Q3. A is a point of the curve B is a point of the line A ((a),((a^2 −a)) ) and B ((b),((b−3)) ) AB^2 = (b−a)^2 +(b−3−a^2 +a)^2 for a given a, AB^2 = f(b) f(b) = (b−a)^2 +(b−3−a^2 +a)^2 f′(b) = 2(b−a)+2(b−3−a^2 +a) f′(b) = 4b−2a^2 −6 f′(b) = 0 ⇔ b = ((a^2 +3)/2) Then : AB^2 = (((a^2 +3)/2)−a)^2 +(((a^2 +3)/2)−3−a^2 +a)^2 AB^2 = ((a^2 /2)−a+(3/2))^2 +(−(a^2 /2)+a−(3/2))^2 AB^2 = 2((a^2 /2)−a+(3/2))^2 AB = (√2)∣(a^2 /2)−a+(3/2)∣ and then : A ((a),((a^2 −a)) ) and B ((((a^2 +3)/2)),(((a^2 −3)/2)) )
$$\mathrm{Q1}. \\ $$$$\mathrm{I}\:=\:\int_{−\mathrm{1}} ^{\mathrm{1}} \mid{x}\mid\left({x}^{\mathrm{3}} +\mathrm{1}\right){dx} \\ $$$$\mathrm{I}\:=\:\int_{−\mathrm{1}} ^{\mathrm{0}} −{x}\left({x}^{\mathrm{3}} +\mathrm{1}\right){dx}+\int_{\mathrm{0}} ^{\mathrm{1}} {x}\left({x}^{\mathrm{3}} +\mathrm{1}\right){dx} \\ $$$$\mathrm{I}\:=\:\left[−\frac{{x}^{\mathrm{5}} }{\mathrm{5}}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right]_{−\mathrm{1}} ^{\mathrm{0}} +\left[\frac{{x}^{\mathrm{5}} }{\mathrm{5}}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\mathrm{I}\:=\:\left(−\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{2}}\right)+\left(\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\mathrm{I}\:=\:\mathrm{1} \\ $$$$\mathrm{Q2}. \\ $$$$\Delta'\:=\:\left(−\mathrm{3}\right)^{\mathrm{2}} −\left(\mathrm{2}\right)\left({k}\right)\:=\:\mathrm{9}−\mathrm{2}{k} \\ $$$$\mathrm{More}\:\mathrm{than}\:\mathrm{one}\:\mathrm{solution}\:\Leftrightarrow\:\Delta'>\mathrm{0} \\ $$$$\Leftrightarrow\:{k}<\frac{\mathrm{9}}{\mathrm{2}} \\ $$$$\Leftrightarrow\:{k}\in\left\{\mathrm{0};\mathrm{1};\mathrm{2};\mathrm{3};\mathrm{4}\right\} \\ $$$$\Rightarrow\:\mathrm{sum}\:=\:\mathrm{0}+\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}\:=\:\mathrm{10} \\ $$$$\mathrm{Q3}. \\ $$$$\mathrm{A}\:\mathrm{is}\:\mathrm{a}\:\mathrm{point}\:\mathrm{of}\:\mathrm{the}\:\mathrm{curve} \\ $$$$\mathrm{B}\:\mathrm{is}\:\mathrm{a}\:\mathrm{point}\:\mathrm{of}\:\mathrm{the}\:\mathrm{line} \\ $$$$\mathrm{A}\begin{pmatrix}{{a}}\\{{a}^{\mathrm{2}} −{a}}\end{pmatrix}\:\mathrm{and}\:\mathrm{B}\begin{pmatrix}{{b}}\\{{b}−\mathrm{3}}\end{pmatrix} \\ $$$$\mathrm{AB}^{\mathrm{2}} \:=\:\left({b}−{a}\right)^{\mathrm{2}} +\left({b}−\mathrm{3}−{a}^{\mathrm{2}} +{a}\right)^{\mathrm{2}} \\ $$$$\mathrm{for}\:\mathrm{a}\:\mathrm{given}\:{a},\:\mathrm{AB}^{\mathrm{2}} \:=\:{f}\left({b}\right) \\ $$$${f}\left({b}\right)\:=\:\left({b}−{a}\right)^{\mathrm{2}} +\left({b}−\mathrm{3}−{a}^{\mathrm{2}} +{a}\right)^{\mathrm{2}} \\ $$$${f}'\left({b}\right)\:=\:\mathrm{2}\left({b}−{a}\right)+\mathrm{2}\left({b}−\mathrm{3}−{a}^{\mathrm{2}} +{a}\right) \\ $$$${f}'\left({b}\right)\:=\:\mathrm{4}{b}−\mathrm{2}{a}^{\mathrm{2}} −\mathrm{6} \\ $$$${f}'\left({b}\right)\:=\:\mathrm{0}\:\Leftrightarrow\:{b}\:=\:\frac{{a}^{\mathrm{2}} +\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{Then}\:: \\ $$$$\mathrm{AB}^{\mathrm{2}} \:=\:\left(\frac{{a}^{\mathrm{2}} +\mathrm{3}}{\mathrm{2}}−{a}\right)^{\mathrm{2}} +\left(\frac{{a}^{\mathrm{2}} +\mathrm{3}}{\mathrm{2}}−\mathrm{3}−{a}^{\mathrm{2}} +{a}\right)^{\mathrm{2}} \\ $$$$\mathrm{AB}^{\mathrm{2}} \:=\:\left(\frac{{a}^{\mathrm{2}} }{\mathrm{2}}−{a}+\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(−\frac{{a}^{\mathrm{2}} }{\mathrm{2}}+{a}−\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\mathrm{AB}^{\mathrm{2}} \:=\:\mathrm{2}\left(\frac{{a}^{\mathrm{2}} }{\mathrm{2}}−{a}+\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\mathrm{AB}\:=\:\sqrt{\mathrm{2}}\mid\frac{{a}^{\mathrm{2}} }{\mathrm{2}}−{a}+\frac{\mathrm{3}}{\mathrm{2}}\mid\:\mathrm{and}\:\mathrm{then}\:: \\ $$$$\mathrm{A}\begin{pmatrix}{{a}}\\{{a}^{\mathrm{2}} −{a}}\end{pmatrix}\:\:\mathrm{and}\:\mathrm{B}\begin{pmatrix}{\frac{{a}^{\mathrm{2}} +\mathrm{3}}{\mathrm{2}}}\\{\frac{{a}^{\mathrm{2}} −\mathrm{3}}{\mathrm{2}}}\end{pmatrix} \\ $$
Commented by abony1303 last updated on 15/Jul/20
no, no it′s ok. :)
$$\left.\mathrm{no},\:\mathrm{no}\:\mathrm{it}'\mathrm{s}\:\mathrm{ok}.\::\right)\: \\ $$
Commented by OlafThorendsen last updated on 15/Jul/20
sorry ! I need to change my glasses ! :−)
$$\mathrm{sorry}\:!\:\mathrm{I}\:\mathrm{need}\:\mathrm{to}\:\mathrm{change}\:\mathrm{my} \\ $$$$\left.\mathrm{glasses}\:!\::−\right) \\ $$
Commented by abony1303 last updated on 15/Jul/20
Thank you ser.Everything is clear, but can you pls explain what is △′ in Q2?
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{ser}.\mathrm{Everything}\:\mathrm{is}\:\mathrm{clear},\:\mathrm{but}\: \\ $$$$\mathrm{can}\:\mathrm{you}\:\mathrm{pls}\:\mathrm{explain}\:\mathrm{what}\:\mathrm{is}\:\bigtriangleup'\:\mathrm{in}\:\mathrm{Q2}? \\ $$
Commented by bemath last updated on 15/Jul/20
Δ = b^2 −4ac discriminant
$$\Delta\:=\:{b}^{\mathrm{2}} −\mathrm{4}{ac}\: \\ $$$${discriminant} \\ $$
Commented by abony1303 last updated on 15/Jul/20
but it′s not the quadratic equation?
$$\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{the}\:\mathrm{quadratic}\:\mathrm{equation}? \\ $$
Commented by OlafThorendsen last updated on 15/Jul/20
Δ′ : reduced discriminant (sorry my english is not fluent) x = ((−b±(√Δ))/(2a)) with Δ = b^2 −4ac but when b is even you can use : x = ((−b′±(√(Δ′)))/a) with b′ = (b/2) and Δ′ = b′^2 −ac (the calculation is faster)
$$\Delta'\::\:\mathrm{reduced}\:\mathrm{discriminant} \\ $$$$\left(\mathrm{sorry}\:\mathrm{my}\:\mathrm{english}\:\mathrm{is}\:\mathrm{not}\:\mathrm{fluent}\right) \\ $$$${x}\:=\:\frac{−{b}\pm\sqrt{\Delta}}{\mathrm{2}{a}}\:\mathrm{with}\:\Delta\:=\:{b}^{\mathrm{2}} −\mathrm{4}{ac} \\ $$$$\mathrm{but}\:\mathrm{when}\:{b}\:\mathrm{is}\:\mathrm{even}\:\mathrm{you}\:\mathrm{can}\:\mathrm{use}\:: \\ $$$${x}\:=\:\frac{−{b}'\pm\sqrt{\Delta'}}{{a}} \\ $$$$\mathrm{with}\:{b}'\:=\:\frac{{b}}{\mathrm{2}}\:\mathrm{and}\:\Delta'\:=\:{b}'^{\mathrm{2}} −{ac} \\ $$$$\left(\mathrm{the}\:\mathrm{calculation}\:\mathrm{is}\:\mathrm{faster}\right) \\ $$
Commented by abony1303 last updated on 15/Jul/20
ser but it′s not the quadratic equation?
$$\mathrm{ser}\:\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{the}\:\mathrm{quadratic}\:\mathrm{equation}? \\ $$

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