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Question Number 198166 by mr W last updated on 12/Oct/23
if f(x)=x^2 +bx+c f(f(1))=f(f(2))=0 and f(1)≠f(2) find f(0)=?
$${if}\:{f}\left({x}\right)={x}^{\mathrm{2}} +{bx}+{c} \\ $$$${f}\left({f}\left(\mathrm{1}\right)\right)={f}\left({f}\left(\mathrm{2}\right)\right)=\mathrm{0}\:{and}\:{f}\left(\mathrm{1}\right)\neq{f}\left(\mathrm{2}\right) \\ $$$${find}\:{f}\left(\mathrm{0}\right)=? \\ $$
Answered by mr W last updated on 12/Oct/23
f(1)=1^2 +b×1+c=b+c+1 f(2)=2^2 +b×2+c=2b+c+4 f(1) and f(2) are the roots of f(x)=0. f(1)+f(2)=−b ⇒b+c+1+2b+c+4=−b ⇒b=−((2c+5)/4) f(1)f(2)=c ⇒(b+c+1)(2b+c+4)=c ⇒(−((2c+5)/4)+c+1)(−((2c+5)/2)+c+4)=c ⇒c=−(3/2) f(0)=c=−(3/2) ✓
$${f}\left(\mathrm{1}\right)=\mathrm{1}^{\mathrm{2}} +{b}×\mathrm{1}+{c}={b}+{c}+\mathrm{1} \\ $$$${f}\left(\mathrm{2}\right)=\mathrm{2}^{\mathrm{2}} +{b}×\mathrm{2}+{c}=\mathrm{2}{b}+{c}+\mathrm{4} \\ $$$${f}\left(\mathrm{1}\right)\:{and}\:{f}\left(\mathrm{2}\right)\:{are}\:{the}\:{roots}\:{of}\:{f}\left({x}\right)=\mathrm{0}. \\ $$$${f}\left(\mathrm{1}\right)+{f}\left(\mathrm{2}\right)=−{b} \\ $$$$\Rightarrow{b}+{c}+\mathrm{1}+\mathrm{2}{b}+{c}+\mathrm{4}=−{b} \\ $$$$\Rightarrow{b}=−\frac{\mathrm{2}{c}+\mathrm{5}}{\mathrm{4}} \\ $$$${f}\left(\mathrm{1}\right){f}\left(\mathrm{2}\right)={c} \\ $$$$\Rightarrow\left({b}+{c}+\mathrm{1}\right)\left(\mathrm{2}{b}+{c}+\mathrm{4}\right)={c} \\ $$$$\Rightarrow\left(−\frac{\mathrm{2}{c}+\mathrm{5}}{\mathrm{4}}+{c}+\mathrm{1}\right)\left(−\frac{\mathrm{2}{c}+\mathrm{5}}{\mathrm{2}}+{c}+\mathrm{4}\right)={c} \\ $$$$\Rightarrow{c}=−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${f}\left(\mathrm{0}\right)={c}=−\frac{\mathrm{3}}{\mathrm{2}}\:\checkmark \\ $$
Answered by Frix last updated on 12/Oct/23
f(x)=x^2 −(x/2)−(3/2) ⇒ f(0)=−(3/2) [Solving this system for a, b, c, d: f(1)=c∧f(2)=d∧f(c)=0∧f(d)=0]
$${f}\left({x}\right)={x}^{\mathrm{2}} −\frac{{x}}{\mathrm{2}}−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$${f}\left(\mathrm{0}\right)=−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$ \\ $$$$\left[\mathrm{Solving}\:\mathrm{this}\:\mathrm{system}\:\mathrm{for}\:{a},\:{b},\:{c},\:{d}:\right. \\ $$$$\left.{f}\left(\mathrm{1}\right)={c}\wedge{f}\left(\mathrm{2}\right)={d}\wedge{f}\left({c}\right)=\mathrm{0}\wedge{f}\left({d}\right)=\mathrm{0}\right] \\ $$
Commented by AST last updated on 12/Oct/23
For f(x)=x^2 −3x+3+_− i; f(1)+f(2)=3=−b;(since f(1) and f(2) are zeros) f(1)+f(2) will not be real(=3)
$${For}\:{f}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{3}\underset{−} {+}{i}; \\ $$$${f}\left(\mathrm{1}\right)+{f}\left(\mathrm{2}\right)=\mathrm{3}=−{b};\left({since}\:{f}\left(\mathrm{1}\right)\:{and}\:{f}\left(\mathrm{2}\right)\:{are}\:{zeros}\right) \\ $$$${f}\left(\mathrm{1}\right)+{f}\left(\mathrm{2}\right)\:{will}\:{not}\:{be}\:{real}\left(=\mathrm{3}\right) \\ $$
Commented by Frix last updated on 12/Oct/23
Thank you, I corrected.
$$\mathrm{Thank}\:\mathrm{you},\:\mathrm{I}\:\mathrm{corrected}. \\ $$
Answered by AST last updated on 12/Oct/23
f(f(1))=f(1+b+c) =1+2b^2 +c^2 +3b+3c+3bc=0 f(1)f(2)=c⇒(1+b+c)(4+2b+c)=c ⇒4+6b+2b^2 +4c+3bc+c^2 =0⇒ (1+2b^2 +c^2 +3b+3c+3bc)+3+3b+c=0⇒3b+c=−3 ⇒f(f(2))=f(4+2b+c)=f(1−b) f(1)+f(2)=−b⇒1+b+c+1−b=2+c=−b ⇒c+b=−2⇒b=−(1/2)⇒c=−(3/2) f(x)=x^2 −(1/2)x−(3/2)⇒f(0)=c=−(3/2)
$${f}\left({f}\left(\mathrm{1}\right)\right)={f}\left(\mathrm{1}+{b}+{c}\right) \\ $$$$=\mathrm{1}+\mathrm{2}{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{3}{b}+\mathrm{3}{c}+\mathrm{3}{bc}=\mathrm{0} \\ $$$${f}\left(\mathrm{1}\right){f}\left(\mathrm{2}\right)={c}\Rightarrow\left(\mathrm{1}+{b}+{c}\right)\left(\mathrm{4}+\mathrm{2}{b}+{c}\right)={c} \\ $$$$\Rightarrow\mathrm{4}+\mathrm{6}{b}+\mathrm{2}{b}^{\mathrm{2}} +\mathrm{4}{c}+\mathrm{3}{bc}+{c}^{\mathrm{2}} =\mathrm{0}\Rightarrow \\ $$$$\left(\mathrm{1}+\mathrm{2}{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{3}{b}+\mathrm{3}{c}+\mathrm{3}{bc}\right)+\mathrm{3}+\mathrm{3}{b}+{c}=\mathrm{0}\Rightarrow\mathrm{3}{b}+{c}=−\mathrm{3} \\ $$$$\Rightarrow{f}\left({f}\left(\mathrm{2}\right)\right)={f}\left(\mathrm{4}+\mathrm{2}{b}+{c}\right)={f}\left(\mathrm{1}−{b}\right) \\ $$$${f}\left(\mathrm{1}\right)+{f}\left(\mathrm{2}\right)=−{b}\Rightarrow\mathrm{1}+{b}+{c}+\mathrm{1}−{b}=\mathrm{2}+{c}=−{b} \\ $$$$\Rightarrow{c}+{b}=−\mathrm{2}\Rightarrow{b}=−\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow{c}=−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${f}\left({x}\right)={x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}{x}−\frac{\mathrm{3}}{\mathrm{2}}\Rightarrow{f}\left(\mathrm{0}\right)={c}=−\frac{\mathrm{3}}{\mathrm{2}} \\ $$

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