Let p(x) be a quadratic polynomial such that distinct reals a and b ,. p(a)=a and p(b)=b then find the roots of. p[p(x)]−x=0Well a and b are surely two of the roots which I found easily but I cannot find the other. I used the approach that P(x)=A(x−a)(x−b)+x then I found p[p(x)]−x from which I can easily see that a and b are it's roots but how can I find the other two.

Question

Let p(x) be a quadratic polynomial such that distinct reals a and b ,. p(a)=a and p(b)=b then find the roots of. p[p(x)]−x=0Well a and b are surely two of the roots which I found easily but I cannot find the other. I used the approach that P(x)=A(x−a)(x−b)+x then I found p[p(x)]−x from which I can easily see that a and b are it&#039;s roots but how can I find the other two.

Cailyn Hanson · Accepted Answer

Note that p(x)=x is a quadratic equation, hence, there exist at most two distinct real numbers that satisfy it. You have p(a)=a and p(b)=b, hence, a,b are these numbers. So, in your equation p(p(x))=x, the only roots are exactly the solutions of the (quadratic) equations p(x)=a and p(x)=b. Since we know that p(a)=a and p(b)=b, a,b are two solutions.Note that, as you have observed,p(x)=C(x−a)(x−b)+x,C≠0. So:p(x)=a⇔C(x−a)(x−b)+x−a=0⇔(x−a)(Cx−Cb+1)=0,so, x=a (we have found that) or x=Cb−1C.Similarly, we can find that the other two roots are x=b or x=Ca−1C.

Let p(x) be a quadratic polynomial such that

Answered question

Answer & Explanation

New Questions in Algebra I