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Question Number 162787 by amin96 last updated on 01/Jan/22

happy new year {a;b;c}∈Z−{0} p(x)=ax^2 +bx+c p(a)=0 p(b)=0 p(1)=?

$$\boldsymbol{\mathrm{happy}}\:\boldsymbol{\mathrm{new}}\:\boldsymbol{\mathrm{year}} \\ $$$$\left\{\boldsymbol{{a}};\boldsymbol{{b}};\boldsymbol{{c}}\right\}\in\mathbb{Z}−\left\{\mathrm{0}\right\} \\ $$$$\boldsymbol{{p}}\left(\boldsymbol{{x}}\right)=\boldsymbol{{ax}}^{\mathrm{2}} +\boldsymbol{{bx}}+\boldsymbol{{c}}\:\:\:\: \\ $$$$\boldsymbol{{p}}\left(\boldsymbol{{a}}\right)=\mathrm{0} \\ $$$$\boldsymbol{{p}}\left(\boldsymbol{{b}}\right)=\mathrm{0} \\ $$$$\boldsymbol{{p}}\left(\mathrm{1}\right)=? \\ $$

Answered by alephzero last updated on 01/Jan/22

p(a) = aa^2 + ba + c = a^3 + ba + c = 0 p(b) = ab^2 + bb + c = ab^2 + b^2 + c = 0 p(1) = a^2 + b + c = ? { ((a^3 + ba + c = 0)),((ab^2 + b^2 + c =0)) :} ⇒ { ((a^3 + ba = −c)),((ab^2 + b^2 = −c)) :} ⇒ a^3 + ba = ab^2 + b^2 ⇒ a(a^2 + b) = b^2 (a + 1) (a_1 ; b_1 ) = (−(1/2); −(1/2)) (a_2 ; b_2 ) = (−2; 4) (a_3 ; b_3 ) = (0; 0) But {a; b; c} ∈ Z−{0} ⇒ (a; b) = (−2; 4) ⇒ { ((−2^3 + (−2) × 4 = −c)),((−2 × 4^2 + 4^2 = −c)) :} { ((−8 + (−8) = −c)),((−36 + 16 = −c)) :} ⇒ −16 = −c ⇒ c = 16 a^2 + b + c = (−2)^2 + 4 + 16 = 4 + 4 + + 16 = 8 + 16 = 24 p(1) = 24

$${p}\left({a}\right)\:=\:{aa}^{\mathrm{2}} \:+\:{ba}\:+\:{c}\:=\:{a}^{\mathrm{3}} \:+\:{ba}\:+\:{c}\:=\:\mathrm{0} \\ $$$${p}\left({b}\right)\:=\:{ab}^{\mathrm{2}} \:+\:{bb}\:+\:{c}\:=\:{ab}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:+\:{c}\:=\:\mathrm{0} \\ $$$${p}\left(\mathrm{1}\right)\:=\:{a}^{\mathrm{2}} \:+\:{b}\:+\:{c}\:=\:? \\ $$$$\begin{cases}{{a}^{\mathrm{3}} +\:{ba}\:+\:{c}\:=\:\mathrm{0}}\\{{ab}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:+\:{c}\:=\mathrm{0}}\end{cases} \\ $$$$\Rightarrow\begin{cases}{{a}^{\mathrm{3}} \:+\:{ba}\:=\:−{c}}\\{{ab}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:=\:−{c}}\end{cases}\: \\ $$$$\Rightarrow\:{a}^{\mathrm{3}} \:+\:{ba}\:=\:{ab}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \\ $$$$\Rightarrow\:{a}\left({a}^{\mathrm{2}} \:+\:{b}\right)\:=\:{b}^{\mathrm{2}} \left({a}\:+\:\mathrm{1}\right) \\ $$$$\left({a}_{\mathrm{1}} ;\:{b}_{\mathrm{1}} \right)\:=\:\left(−\frac{\mathrm{1}}{\mathrm{2}};\:−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\left({a}_{\mathrm{2}} ;\:{b}_{\mathrm{2}} \right)\:=\:\left(−\mathrm{2};\:\mathrm{4}\right) \\ $$$$\left({a}_{\mathrm{3}} ;\:{b}_{\mathrm{3}} \right)\:=\:\left(\mathrm{0};\:\mathrm{0}\right) \\ $$$$\mathrm{But}\:\left\{{a};\:{b};\:{c}\right\}\:\in\:\mathbb{Z}−\left\{\mathrm{0}\right\} \\ $$$$\Rightarrow\:\left({a};\:{b}\right)\:=\:\left(−\mathrm{2};\:\mathrm{4}\right) \\ $$$$\Rightarrow\begin{cases}{−\mathrm{2}^{\mathrm{3}} \:+\:\left(−\mathrm{2}\right)\:×\:\mathrm{4}\:=\:−{c}}\\{−\mathrm{2}\:×\:\mathrm{4}^{\mathrm{2}} \:+\:\mathrm{4}^{\mathrm{2}} \:=\:−{c}}\end{cases} \\ $$$$\begin{cases}{−\mathrm{8}\:+\:\left(−\mathrm{8}\right)\:=\:−{c}}\\{−\mathrm{36}\:+\:\mathrm{16}\:=\:−{c}}\end{cases} \\ $$$$\Rightarrow\:−\mathrm{16}\:=\:−{c} \\ $$$$\Rightarrow\:{c}\:=\:\mathrm{16} \\ $$$${a}^{\mathrm{2}} \:+\:{b}\:+\:{c}\:=\:\left(−\mathrm{2}\right)^{\mathrm{2}} \:+\:\mathrm{4}\:+\:\mathrm{16}\:=\:\mathrm{4}\:+\:\mathrm{4}\:+ \\ $$$$+\:\mathrm{16}\:=\:\mathrm{8}\:+\:\mathrm{16}\:=\:\mathrm{24} \\ $$$${p}\left(\mathrm{1}\right)\:=\:\mathrm{24} \\ $$

Commented by mr W last updated on 01/Jan/22

solution is not unique. f(x)=x^2 +x−2 is also ok. f(1)=0. in fact any a=b=n, c=−n^2 (n+1) with n∈Z is ok.

$${solution}\:{is}\:{not}\:{unique}. \\ $$$${f}\left({x}\right)={x}^{\mathrm{2}} +{x}−\mathrm{2}\:{is}\:{also}\:{ok}. \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{0}. \\ $$$${in}\:{fact}\:{any}\:{a}={b}={n},\:{c}=−{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)\:{with} \\ $$$${n}\in{Z}\:{is}\:{ok}. \\ $$

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