Let-p-x-be-a-quadratic-polynomial-such-that-for-distinct-and-p-and-p-prove-that-and-are-roots-of-p-p-x-x-0-Find-the-remaining-roots-

Question Number 61269 by alphaprime last updated on 31/May/19

$$Letp\left(x\right)beaquadraticpolynomialsuch$$$$thatfordistinct\alpha and\beta ,$$$$p\left(\alpha \right)=\alpha andp\left(\beta \right)=\beta$$$$provethat\alpha and\beta arerootsofp\left[p\left(x\right)\right]-x=0$$$$Findtheremainingroots.$$

Answered by ajfour last updated on 31/May/19

$$p\left(x\right)=a{(x-h)}^2+k$$$$\alpha ,\beta arerootsof$$$$a{(r-h)}^2-r+k=0\dots \left(i\right)$$$$p\left[p\left(x\right)\right]-x=a{\{a{(x-h)}^2+k-h\}}^2+k-x$$$$Andsincefrom\left(i\right)$$$$a{(r-h)}^2+k=r\dots .\left(ii\right)$$$$p\left[p\left(r\right)\right]-r=a{\{a{(x-r)}^2+k-h\}}^2-r+k$$$$using\left(ii\right)weget$$$$p\left[p\left(r\right)\right]-r=a{\{r-h\}}^2-r+k$$$$=0using\left(i\right)$$$$hence\alpha ,\beta arerootsofp\left[p\left(x\right)\right]-x=0.$$$$Again$$$$p\left[p\left(x\right)\right]-x=a{\{a{(x-h)}^2+k-h\}}^2+k-x$$$$or=a^3{(x-h)}^4+2a^2(k-h){(x-h)}^2$$$$+a{(k-h)}^2+k-x$$$$=a^3{x}^4-4a^3{hx}^3+[6a^3{h}^2+2a^2(k-h)]x^2$$$$-[4a^3{h}^3-+4a^{2}h(k-h)+1]x$$$$+a^3{h}^4+2a^2(k-h)h^2+a{(k-h)}^2+k$$$$Alsoletsassume$$$$p\left[p\left(x\right)\right]-x=a^3[x^2-(2ah+1)x+h^2+\frac{k}{a}](x-\gamma )(x-\delta )$$$$=a^3\{x^4-(\gamma +\delta +2ah+1)x^3$$$$+[\gamma \delta +(2ah+1)(\gamma +\delta )+h^2+\frac{k}{a}]x^2$$$$+(h^2+\frac{k}{a})\gamma \delta \}$$$$comparingthetwoexpressions$$$$forp\left[p\left(x\right)\right]-xweconclude$$$$\gamma +\delta =4h-2ah-1$$$$\gamma \delta (h^2+\frac{k}{a})=a^3{h}^4+2a^2(k-h)h^2+a{(k-h)}^2+k$$$$butfrom\left(i\right)$$$$2h+\frac{1}{a}=\alpha +\beta$$$$h^2+\frac{k}{a}=\alpha \beta$$$$\dots ..$$

Answered by perlman last updated on 31/May/19

$$p\left(x\right)=(x-a)(x-b)+x$$$$p\left[p\left(x\right)\right]=(p\left(x\right)-a)(p\left(x\right)-b)+p\left(x\right)$$$$pp\left(x\right)-x=0$$$$==>(p\left(x\right)-a)(p\left(x\right)-b)+p\left(x\right)-x=0$$$$=((x-a)(x-b)+x-a)((x-a)(x-b)+x-b)+(x-a)(x-b)=0$$$$(x-a)(x-b+1)(x-b)(x-a+1)+(x-a)(x-b)=0$$$$(x-a)(x-b)[(x-b+1)(x-a+1)+1]=0$$$$letsolve(x-b+1)(x-a+1)+1=0$$$$x^2-(b+a-2)x+(a-1)(b-1)+1=0$$$$\mathrm{\Delta }={(b+a-2)}^2-4(ab-a-b+2)=(b^2+a^2+4+2ba-4a-4b)-4ab+4a+4b-8$$$$=b^2+a^2-2ab-4=(b-a-2)(b-a+2)$$$$\text{You can't use 'macro parameter character \#' in math mode}$$$$===>\mathrm{\Delta }⩾0<==>b⩾a+2$$$$ifb⩾a+2$$$$X_{1.2}=\frac{(b+a-2)\underset{-}{+}\sqrt{(b-a-2)(b-a+2)}}{2}$$$$ifa<b<=a+2$$$$X_{1.2}=\frac{(b+a-2)\underset{-}{+}i\sqrt{(a+2-b)(b-a+2)}}{2}$$$$solutionareifb\in ]a+2;+\mathrm{\infty }[solutionsare$$$$\{b,a,\frac{(b+a-2)\underset{-}{+}\sqrt{(b-a-2)(b-a+2)}}{2}\}$$$$ifb\in ]a;a+2[solutionsare$$$$\{b,a,\frac{(b+a-2)\underset{-}{+}i\sqrt{(b-a-2)(b-a+2)}}{2}\}$$$$$$

Commented by alphaprime last updated on 31/May/19

Tanmay sir generalized and solution is pretty nice , but yours is valid in entire domain , But comment something on both the solutions , any remarks you wanna add , please ...

Commented by perlman last updated on 31/May/19

$$itackp\left(x\right)=(x-a)(x-b)+x$$$$butgeneralyimusttackp=s(x-a)(x-b)+x$$$$withesconstant$$

Commented by alphaprime last updated on 31/May/19

You mentioned it finally , In Tanmay sir's solution It's dependent on the parameter A you've left , which you haven't taken as arbitrary constant, so does it affect his or your solution or neither ones ?! it's amazing

Answered by tanmay last updated on 31/May/19

$$p\left(x\right)=A(x-\alpha )(x-\beta )+x$$$$p\left[p\left(x\right)\right]$$$$=A[A(x-\alpha )(x-\beta )+x-\alpha ](A(x-\alpha )(x-\beta )+x-\beta ]+$$$$A(x-\alpha )(x-\beta )+x-x$$$$so$$$$p\left[p\left(\alpha \right)\right]=0$$$$p\left[p\left(\beta \right)\right]=0$$$$nxtpart\dots$$$$p\left[p\left(x\right)\right]$$$$=A\left[(x-\alpha )\{A(x-\beta )+1\}\right]\left[(x-\beta )\{A(x-\alpha )+1\}\right]+if$$$$A(x-\alpha )(x-\beta )$$$$forothertworootp\left[p\left(x\right)\right]=0$$$$x-\alpha =mx-\beta =n$$$$A\left[m(An+1)\right][n(Am+1]+Amn=0$$$$(Amn+m)(Amn+n)+mn=0$$$$A^2{\left(mn\right)}^2+Amn(m+n)+mn+mn=0$$$$mn[A^2\left(mn\right)+A(m+n)+2]=0$$$$when{A}^2\left(mn\right)+A(m+n)+2=0$$$$A^2(x-\alpha )(x-\beta )+A(x-\alpha +x-\beta )+2=0$$$$A^2(x^2-x\alpha -x\beta +\alpha \beta )+2Ax-A(\alpha +\beta )+2=0$$$$x^2\left(A^2\right)+x(2A-A^2\alpha -A^2\beta )+A^2\alpha \beta -A(\alpha +\beta )+2=0$$$${(Ax+1)}^2+x(-A^2\alpha -A^2\beta )+(A^2\alpha \beta -A\alpha -A\beta +1)=0$$$${(Ax+1)}^2-A^{2}x(\alpha +\beta )+A\alpha (A\beta -1)-(A\beta -1)=0$$$${(Ax+1)}^2-A^{2}x(\alpha +\beta )+(A\alpha -1)(A\beta -1)=0$$$$A^2{x}^2+2Ax+1-A^{2}x(\alpha +\beta )+(A\alpha -1)(A\beta -1)=0$$$$A^2{x}^2-Ax(\alpha +\beta -2)+(A\alpha -1)(A\beta -1)+1=0$$$$x=\frac{A(\alpha +\beta -2)\pm \sqrt{A^2{(\alpha +\beta -2)}^2-4A^2[(A\alpha -1)(A\beta -1)+1]}}{2A^2}$$$$x=\frac{\alpha +\beta -2\pm \sqrt{{(\alpha +\beta -2)}^2-4[(A\alpha -1)(A\beta -1)+1}}{2A}$$$$$$$$$$$$$$$$$$

Commented by alphaprime last updated on 31/May/19

Very Nice cognizant

Commented by tanmay last updated on 31/May/19

$$thankyousir\dots$$

Commented by alphaprime last updated on 31/May/19

Question no.61211 , I seriously need your help sir , I can't even come on telegram due to issues, I'm seeking intense help , I don't expect much but there's a hope , Mjs sir is with me in the workspace and has promised to provide solution till Monday, I need you and other intellectuals to join him and resolve the system Indeed help me out please sir ��

Commented by ajfour last updated on 31/May/19

$$Wetooaregladtohaveyouhere$$$$ontheforumSir.Yourquestions$$$$areprecious.$$

Commented by alphaprime last updated on 31/May/19

Sir I'm welcoming you in my workspace and please don't disappoint me , I'll love to notice presence of higher intellectual faculty , Just provide me your email ��

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