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Question Number 44543 by ajfour last updated on 01/Oct/18

If y =f(x) = ax^2 +bx+c and at some x, say x= p ∫_0 ^( p) ydx = y(p)= y ′(p) = y ′′(p)= p , then find p .

$${If}\:\:{y}\:={f}\left({x}\right)\:=\:{ax}^{\mathrm{2}} +{bx}+{c} \\ $$$${and}\:\:{at}\:{some}\:{x},\:{say}\:\:{x}=\:{p} \\ $$$$\int_{\mathrm{0}} ^{\:\:{p}} {ydx}\:=\:{y}\left({p}\right)=\:{y}\:'\left({p}\right)\:=\:{y}\:''\left({p}\right)=\:{p}\:, \\ $$$${then}\:{find}\:\boldsymbol{{p}}\:. \\ $$

Commented by MrW3 last updated on 01/Oct/18

y′(x)=2ax+b y′′(x)=2a 2a=p⇒a=(p/2) 2×(p/2)×p+b=p⇒b=p(1−p) (p/2)×p^2 +p(1−p)×p+c=p ⇒c=p−p^2 +(p^3 /2) ∫_0 ^p ydx=(a/3)×p^3 +(b/2)×p^2 +cp=p (a/3)×p^2 +(b/2)×p+c=1 (p/6)×p^2 +((p(1−p))/2)×p+p−p^2 +(p^3 /2)=1 (p^3 /6)+(p^2 /2)−(p^3 /2)+p−p^2 +(p^3 /2)=1 (p^3 /6)−(p^2 /2)+p=1 p^3 −3p^2 +6p−6=0 ⇒p≈1.6

$${y}'\left({x}\right)=\mathrm{2}{ax}+{b} \\ $$$${y}''\left({x}\right)=\mathrm{2}{a} \\ $$$$\mathrm{2}{a}={p}\Rightarrow{a}=\frac{{p}}{\mathrm{2}} \\ $$$$\mathrm{2}×\frac{{p}}{\mathrm{2}}×{p}+{b}={p}\Rightarrow{b}={p}\left(\mathrm{1}−{p}\right) \\ $$$$\frac{{p}}{\mathrm{2}}×{p}^{\mathrm{2}} +{p}\left(\mathrm{1}−{p}\right)×{p}+{c}={p} \\ $$$$\Rightarrow{c}={p}−{p}^{\mathrm{2}} +\frac{{p}^{\mathrm{3}} }{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{{p}} {ydx}=\frac{{a}}{\mathrm{3}}×{p}^{\mathrm{3}} +\frac{{b}}{\mathrm{2}}×{p}^{\mathrm{2}} +{cp}={p} \\ $$$$\frac{{a}}{\mathrm{3}}×{p}^{\mathrm{2}} +\frac{{b}}{\mathrm{2}}×{p}+{c}=\mathrm{1} \\ $$$$\frac{{p}}{\mathrm{6}}×{p}^{\mathrm{2}} +\frac{{p}\left(\mathrm{1}−{p}\right)}{\mathrm{2}}×{p}+{p}−{p}^{\mathrm{2}} +\frac{{p}^{\mathrm{3}} }{\mathrm{2}}=\mathrm{1} \\ $$$$\frac{{p}^{\mathrm{3}} }{\mathrm{6}}+\frac{{p}^{\mathrm{2}} }{\mathrm{2}}−\frac{{p}^{\mathrm{3}} }{\mathrm{2}}+{p}−{p}^{\mathrm{2}} +\frac{{p}^{\mathrm{3}} }{\mathrm{2}}=\mathrm{1} \\ $$$$\frac{{p}^{\mathrm{3}} }{\mathrm{6}}−\frac{{p}^{\mathrm{2}} }{\mathrm{2}}+{p}=\mathrm{1} \\ $$$${p}^{\mathrm{3}} −\mathrm{3}{p}^{\mathrm{2}} +\mathrm{6}{p}−\mathrm{6}=\mathrm{0} \\ $$$$\Rightarrow{p}\approx\mathrm{1}.\mathrm{6} \\ $$

Commented by MrW3 last updated on 01/Oct/18

thanks for checking sir!

$${thanks}\:{for}\:{checking}\:{sir}! \\ $$

Commented by ajfour last updated on 01/Oct/18

yes sir, it is correct there, i dint notice properly, please pardon. Thank you.

$${yes}\:{sir},\:{it}\:{is}\:{correct}\:{there},\:{i}\:{dint} \\ $$$${notice}\:{properly},\:{please}\:{pardon}. \\ $$$${Thank}\:{you}. \\ $$

Answered by ajfour last updated on 01/Oct/18

p=2a=2ap+b=ap^2 +bp+c = ((ap^3 )/3)+((bp^2 )/2)+cp ⇒ ap^2 +bp+c = p ..(i) ⇒ ((ap^2 )/3)+((bp)/2)+c = 1 ..(ii) (i)−(ii) gives ((2ap^2 )/3)+((bp)/2) = p−1 , and we already have 2ap+b = p with 2a=p , b=p−p^2 So (p^3 /3)+(p/2)(p−p^2 )=p−1 ⇒ (p^3 /6)−(p^2 /2)+p−1 = 0 or p^3 −3p^2 +6p−6 =0 let p = t+1 ⇒ t^3 +3t+1−6t−3+6t = 0 ⇒ t^3 +3t−2 =0 let t=u+v ⇒ u^3 +v^3 +3(u+v)(uv+1)= 2 further let uv=−1 then u^3 +v^3 = 2 u^3 , v^3 are roots of quadratic z^2 −2z−1= 0 z = 1±(√2) t= u+v = ((√2)+1)^(1/3) −((√2)−1)^(1/3) p = t+1 ≈ 1.59607164

$${p}=\mathrm{2}{a}=\mathrm{2}{ap}+{b}={ap}^{\mathrm{2}} +{bp}+{c} \\ $$$$\:\:\:\:\:=\:\frac{{ap}^{\mathrm{3}} }{\mathrm{3}}+\frac{{bp}^{\mathrm{2}} }{\mathrm{2}}+{cp} \\ $$$$\Rightarrow\:\:{ap}^{\mathrm{2}} +{bp}+{c}\:=\:{p}\:\:\:\:..\left({i}\right) \\ $$$$\Rightarrow\:\frac{{ap}^{\mathrm{2}} }{\mathrm{3}}+\frac{{bp}}{\mathrm{2}}+{c}\:=\:\mathrm{1}\:\:\:\:\:..\left({ii}\right) \\ $$$$\left({i}\right)−\left({ii}\right)\:{gives} \\ $$$$\frac{\mathrm{2}{ap}^{\mathrm{2}} }{\mathrm{3}}+\frac{{bp}}{\mathrm{2}}\:=\:{p}−\mathrm{1}\:\:,\:{and}\:{we}\:{already} \\ $$$${have}\:\:\:\:\mathrm{2}{ap}+{b}\:=\:{p} \\ $$$${with}\:\mathrm{2}{a}={p}\:,\:\:\:{b}={p}−{p}^{\mathrm{2}} \\ $$$${So}\:\:\:\:\frac{{p}^{\mathrm{3}} }{\mathrm{3}}+\frac{{p}}{\mathrm{2}}\left({p}−{p}^{\mathrm{2}} \right)={p}−\mathrm{1} \\ $$$$\Rightarrow\:\:\frac{\boldsymbol{{p}}^{\mathrm{3}} }{\mathrm{6}}−\frac{\boldsymbol{{p}}^{\mathrm{2}} }{\mathrm{2}}+\boldsymbol{{p}}−\mathrm{1}\:=\:\mathrm{0} \\ $$$${or}\:\:\:{p}^{\mathrm{3}} −\mathrm{3}{p}^{\mathrm{2}} +\mathrm{6}{p}−\mathrm{6}\:=\mathrm{0} \\ $$$${let}\:{p}\:=\:{t}+\mathrm{1} \\ $$$$\Rightarrow\:\:{t}^{\mathrm{3}} +\mathrm{3}{t}+\mathrm{1}−\mathrm{6}{t}−\mathrm{3}+\mathrm{6}{t}\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\:{t}^{\mathrm{3}} +\mathrm{3}{t}−\mathrm{2}\:=\mathrm{0} \\ $$$${let}\:\:{t}={u}+{v} \\ $$$$\Rightarrow\:\:{u}^{\mathrm{3}} +{v}^{\mathrm{3}} +\mathrm{3}\left({u}+{v}\right)\left({uv}+\mathrm{1}\right)=\:\mathrm{2} \\ $$$${further}\:{let}\:{uv}=−\mathrm{1} \\ $$$${then}\:\:\:{u}^{\mathrm{3}} +{v}^{\mathrm{3}} \:=\:\mathrm{2} \\ $$$${u}^{\mathrm{3}} ,\:{v}^{\mathrm{3}} \:{are}\:{roots}\:{of}\:{quadratic} \\ $$$$\:\:\:\:\:\boldsymbol{{z}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{{z}}−\mathrm{1}=\:\mathrm{0} \\ $$$$\:\:\:\boldsymbol{{z}}\:=\:\mathrm{1}\pm\sqrt{\mathrm{2}} \\ $$$$\:\:{t}=\:{u}+{v}\:=\:\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{1}/\mathrm{3}} −\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$$\:\:\boldsymbol{{p}}\:=\:{t}+\mathrm{1}\:\approx\:\mathrm{1}.\mathrm{59607164} \\ $$

Commented by MrW3 last updated on 01/Oct/18

I knew you′ll give the exact solution, thanks for the perfect result!

$${I}\:{knew}\:{you}'{ll}\:{give}\:{the}\:{exact}\:{solution}, \\ $$$${thanks}\:{for}\:{the}\:{perfect}\:{result}! \\ $$

Commented by ajfour last updated on 01/Oct/18

thanks mrW Sir (for Cardano′s).

$${thanks}\:{mrW}\:{Sir}\:\left({for}\:{Cardano}'{s}\right). \\ $$

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