Find all integer values of m such that the equation 9−x=3m−x2+9x−x has exactly four distinct real roots.

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Dixie Reed · Accepted Answer

Step 1We have x+9−x=3m−x2+9x,which is for 0≤x≤9 and 3m−x2+9x≥0 is equivalent to9+2x(9−x)−x(9−x)=3mor 10−(1−x(9−x))2=3m,which gives 3m≤10 and m≤3.Step 2Also, since x+9−x=9+2x(9−x)≥3,we obtain:3m−x2+9x≥9.Thus, 3m≥x2−9x+9=(x−4.5)2−11.25≥−11.25,which gives −3≤m≤3.Now, consider f(x)=10−(1−x(9−x))2.We see that f has two maximum points for x(9−x)=1 and the minimum point for x=4.5.Step 3Thus, our equation has four distinct roots forf(0)≤3m&amp;lt;f(xmax)or 9≤3m&amp;lt;10,which gives m=3.

zevillageobau · Accepted Answer

Explanation:Step 19−x=3m−x2+9x−xEquations 9−x=x has no solution. For the moment we assume that equation 3m−x2+9x=x has also no solution. Therefore the roots of equation 9−x=3m−x2+9x must be satisfied by x; considering x∈[0,9], we may write for integer x:x=0.1,4,99−x=3m−x2+9xm=x2−10x+93Step 2Which gives:(x,m)=(0,3),(1,0),(4,−5),(9,0)Among these only (x,m)=(0,3) satisfies the original equation.Moreover for m=3 we also get x=9 in equation.If we assume that equation 3m−x2+9x=x has solution then it must satisfy the original equation, I checked; it does not.

Find all integer values of m such that

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