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Question Number 198166 by mr W last updated on 12/Oct/23

$$iff\left(x\right)=x^2+bx+c$$$$f\left(f\left(1\right)\right)=f\left(f\left(2\right)\right)=0andf\left(1\right)\ne f\left(2\right)$$$$findf\left(0\right)=?$$

Answered by mr W last updated on 12/Oct/23

$$f\left(1\right)=1^2+b\times 1+c=b+c+1$$$$f\left(2\right)=2^2+b\times 2+c=2b+c+4$$$$f\left(1\right)andf\left(2\right)aretherootsoff\left(x\right)=0.$$$$f\left(1\right)+f\left(2\right)=-b$$$$\Rightarrow b+c+1+2b+c+4=-b$$$$\Rightarrow b=-\frac{2c+5}{4}$$$$f\left(1\right)f\left(2\right)=c$$$$\Rightarrow (b+c+1)(2b+c+4)=c$$$$\Rightarrow (-\frac{2c+5}{4}+c+1)(-\frac{2c+5}{2}+c+4)=c$$$$\Rightarrow c=-\frac{3}{2}$$$$f\left(0\right)=c=-\frac{3}{2}✓$$

Answered by Frix last updated on 12/Oct/23

$$f\left(x\right)=x^2-\frac{x}{2}-\frac{3}{2}$$$$\Rightarrow$$$$f\left(0\right)=-\frac{3}{2}$$$$$$$$[\mathrm{Solving}\mathrm{this}\mathrm{system}\mathrm{for}a,b,c,d:$$$$f\left(1\right)=c\wedge f\left(2\right)=d\wedge f\left(c\right)=0\wedge f\left(d\right)=0]$$

Commented by deleteduser1 last updated on 12/Oct/23

$$Forf\left(x\right)=x^2-3x+3\underset{-}{+}i;$$$$f\left(1\right)+f\left(2\right)=3=-b;\left(sincef\left(1\right)andf\left(2\right)arezeros\right)$$$$f\left(1\right)+f\left(2\right)willnotbereal(=3)$$

Commented by Frix last updated on 12/Oct/23

$$\mathrm{Thank}\mathrm{you},\mathrm{I}\mathrm{corrected}.$$

Answered by deleteduser1 last updated on 12/Oct/23

$$f\left(f\left(1\right)\right)=f(1+b+c)$$$$=1+2b^2+c^2+3b+3c+3bc=0$$$$f\left(1\right)f\left(2\right)=c\Rightarrow (1+b+c)(4+2b+c)=c$$$$\Rightarrow 4+6b+2b^2+4c+3bc+c^2=0\Rightarrow$$$$(1+2b^2+c^2+3b+3c+3bc)+3+3b+c=0\Rightarrow 3b+c=-3$$$$\Rightarrow f\left(f\left(2\right)\right)=f(4+2b+c)=f(1-b)$$$$f\left(1\right)+f\left(2\right)=-b\Rightarrow 1+b+c+1-b=2+c=-b$$$$\Rightarrow c+b=-2\Rightarrow b=-\frac{1}{2}\Rightarrow c=-\frac{3}{2}$$$$f\left(x\right)=x^2-\frac{1}{2}x-\frac{3}{2}\Rightarrow f\left(0\right)=c=-\frac{3}{2}$$

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