Given-the-roots-of-the-quadratic-equation-4x-2-4x-5-0-are-and-f-x-is-a-quadratic-function-where-f-f-and-f-0-6-Find-f-x-

Question Number 163323 by ZiYangLee last updated on 06/Jan/22

Given the roots of the quadratic equation 4x^2 −4x+5=0 are α and β. f(x) is a quadratic function where f(α)=β , f(β)=α and f(0)=6 . Find f(x).

$$\mathrm{Given}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{quadratic}\:\mathrm{equation} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{5}=\mathrm{0}\:\mathrm{are}\:\alpha\:\mathrm{and}\:\beta.\:\: \\ $$$${f}\left({x}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{quadratic}\:\mathrm{function}\:\mathrm{where}\:{f}\left(\alpha\right)=\beta\:, \\ $$$${f}\left(\beta\right)=\alpha\:\mathrm{and}\:{f}\left(\mathrm{0}\right)=\mathrm{6}\:.\: \\ $$$$\mathrm{Find}\:{f}\left({x}\right). \\ $$

Commented by cortano1 last updated on 06/Jan/22

let f(x)= px^2 +qx+6 ⇒ { ((f(α)=pα^2 +qα+6=β)),((f(β)=pβ^2 +qβ+6=α)) :} (1)+(2)⇒p[(α+β)^2 −2αβ ]+q(α+β)+12=α+β ⇒p[ 1−(5/2) ]+q(1)+12=1 ⇒−(3/2)p+q=−11 ⇒−3p+2q=−22 ⇒q=((3p−22)/2) ∴ f(x)= px^2 +(((3p−22)/2))x+6

$${let}\:{f}\left({x}\right)=\:{px}^{\mathrm{2}} +{qx}+\mathrm{6}\: \\ $$$$\Rightarrow\begin{cases}{{f}\left(\alpha\right)={p}\alpha^{\mathrm{2}} +{q}\alpha+\mathrm{6}=\beta}\\{{f}\left(\beta\right)={p}\beta^{\mathrm{2}} +{q}\beta+\mathrm{6}=\alpha}\end{cases} \\ $$$$\:\left(\mathrm{1}\right)+\left(\mathrm{2}\right)\Rightarrow{p}\left[\left(\alpha+\beta\right)^{\mathrm{2}} −\mathrm{2}\alpha\beta\:\right]+{q}\left(\alpha+\beta\right)+\mathrm{12}=\alpha+\beta \\ $$$$\Rightarrow{p}\left[\:\mathrm{1}−\frac{\mathrm{5}}{\mathrm{2}}\:\right]+{q}\left(\mathrm{1}\right)+\mathrm{12}=\mathrm{1} \\ $$$$\Rightarrow−\frac{\mathrm{3}}{\mathrm{2}}{p}+{q}=−\mathrm{11} \\ $$$$\Rightarrow−\mathrm{3}{p}+\mathrm{2}{q}=−\mathrm{22} \\ $$$$\Rightarrow{q}=\frac{\mathrm{3}{p}−\mathrm{22}}{\mathrm{2}} \\ $$$$\therefore\:{f}\left({x}\right)=\:{px}^{\mathrm{2}} +\left(\frac{\mathrm{3}{p}−\mathrm{22}}{\mathrm{2}}\right){x}+\mathrm{6}\: \\ $$

Answered by Rasheed.Sindhi last updated on 06/Jan/22

f(x)=ax^2 +bx+c (say) f(0)=c=6 f(α)=aα^2 +bα+6=β.............(i) f(β)=aβ^2 +bβ+6=α..............(ii) (i)+(ii): a(α^2 +β^2 )+b(α+β)+12=α+β a{(α+β)^2 −2αβ}+b(α+β)+12=α+β 4x^2 −4x+5=0⇒ { ((α+β=1)),((αβ=(5/4))) :} a{(1)^2 −2((5/4))}+b(1)+12=1 a(−(3/2))+b=−11 3a−2b=22.....................A (i)−(ii): f(α)−f(β)=a(α^2 −β^2 )+b(α−β)=−(α−β) =a(α+β)+b=−(α+β) ;α≠β a(1)+b=−(1) a+b=−1......................B A & B: 3a−2(−1−a)=22 3a+2+2a=22 a=4⇒b=−5 f(x)=4x^2 −5x+6

$${f}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+{c}\:\left({say}\right) \\ $$$${f}\left(\mathrm{0}\right)={c}=\mathrm{6} \\ $$$${f}\left(\alpha\right)={a}\alpha^{\mathrm{2}} +{b}\alpha+\mathrm{6}=\beta………….\left({i}\right) \\ $$$${f}\left(\beta\right)={a}\beta^{\mathrm{2}} +{b}\beta+\mathrm{6}=\alpha…………..\left({ii}\right) \\ $$$$\left({i}\right)+\left({ii}\right): \\ $$$${a}\left(\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} \right)+{b}\left(\alpha+\beta\right)+\mathrm{12}=\alpha+\beta \\ $$$${a}\left\{\left(\alpha+\beta\right)^{\mathrm{2}} −\mathrm{2}\alpha\beta\right\}+{b}\left(\alpha+\beta\right)+\mathrm{12}=\alpha+\beta \\ $$$$\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{5}=\mathrm{0}\Rightarrow\begin{cases}{\alpha+\beta=\mathrm{1}}\\{\alpha\beta=\frac{\mathrm{5}}{\mathrm{4}}}\end{cases} \\ $$$${a}\left\{\left(\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}\left(\frac{\mathrm{5}}{\mathrm{4}}\right)\right\}+{b}\left(\mathrm{1}\right)+\mathrm{12}=\mathrm{1} \\ $$$${a}\left(−\frac{\mathrm{3}}{\mathrm{2}}\right)+{b}=−\mathrm{11} \\ $$$$\mathrm{3}{a}−\mathrm{2}{b}=\mathrm{22}…………………{A} \\ $$$$\left({i}\right)−\left({ii}\right): \\ $$$${f}\left(\alpha\right)−{f}\left(\beta\right)={a}\left(\alpha^{\mathrm{2}} −\beta^{\mathrm{2}} \right)+{b}\left(\alpha−\beta\right)=−\left(\alpha−\beta\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={a}\left(\alpha+\beta\right)+{b}=−\left(\alpha+\beta\right)\:\:;\alpha\neq\beta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{a}\left(\mathrm{1}\right)+{b}=−\left(\mathrm{1}\right) \\ $$$${a}+{b}=−\mathrm{1}………………….{B} \\ $$$${A}\:\&\:{B}: \\ $$$$\mathrm{3}{a}−\mathrm{2}\left(−\mathrm{1}−{a}\right)=\mathrm{22} \\ $$$$\mathrm{3}{a}+\mathrm{2}+\mathrm{2}{a}=\mathrm{22} \\ $$$${a}=\mathrm{4}\Rightarrow{b}=−\mathrm{5} \\ $$$${f}\left({x}\right)=\mathrm{4}{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{6} \\ $$

Commented by peter frank last updated on 06/Jan/22