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If-and-are-the-roots-of-equation-x-2-px-q-0-and-2-2-are-roots-of-the-equation-x-2-rx-s-0-show-that-the-equation-x-2-4qx-2q-2-r-0-has-real-roots-

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Question Number 19945 by Tinkutara last updated on 18/Aug/17
If α and β are the roots of equation x^2 + px + q = 0 and α^2 , β^2 are roots of the equation x^2 − rx + s = 0, show that the equation x^2 − 4qx + 2q^2 − r = 0 has real roots.
$$\mathrm{If}\:\alpha\:\mathrm{and}\:\beta\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{equation} \\ $$$${x}^{\mathrm{2}} \:+\:{px}\:+\:{q}\:=\:\mathrm{0}\:\mathrm{and}\:\alpha^{\mathrm{2}} ,\:\beta^{\mathrm{2}} \:\mathrm{are}\:\mathrm{roots}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{equation}\:{x}^{\mathrm{2}} \:−\:{rx}\:+\:{s}\:=\:\mathrm{0},\:\mathrm{show} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{equation}\:{x}^{\mathrm{2}} \:−\:\mathrm{4}{qx}\:+\:\mathrm{2}{q}^{\mathrm{2}} \:−\:{r}\:=\:\mathrm{0} \\ $$$$\mathrm{has}\:\mathrm{real}\:\mathrm{roots}. \\ $$
Answered by Rasheed.Sindhi last updated on 19/Aug/17
x^2 +px+q=0, roots:α,β (given) x^2 −rx+s=0, roots:α^2 ,β^2 (given) x^2 −4qx+2q^2 −r=0,roots are real.(to be proved) −−−−−−−−−−− α+β=−p ,αβ=q α^2 +β^2 =r,α^2 β^2 =s=q^2 x^2 −4qx+2q^2 −r=0 ⇒x^2 −4(αβ)x+2(αβ)^2 −(α^2 +β^2 )=0 x=((4αβ±(√(16α^2 β^2 −4(1)(2(αβ)^2 −(α^2 +β^2 )))))/2) x=((4αβ±2(√(4α^2 β^2 −2(αβ)^2 +(α^2 +β^2 ))))/2) =2αβ±(√(4α^2 β^2 −2α^2 β^2 +α^2 +β^2 )) =2αβ±(√(2α^2 β^2 +α^2 +β^2 )) α^2 ,β^2 ≥0⇒2α^2 β^2 +α^2 +β^2 ≥0 Hence the roots are real.
$$\mathrm{x}^{\mathrm{2}} +\mathrm{px}+\mathrm{q}=\mathrm{0},\:\mathrm{roots}:\alpha,\beta\:\left(\mathrm{given}\right) \\ $$$$\mathrm{x}^{\mathrm{2}} −\mathrm{rx}+\mathrm{s}=\mathrm{0},\:\mathrm{roots}:\alpha^{\mathrm{2}} ,\beta^{\mathrm{2}} \left(\mathrm{given}\right) \\ $$$$\mathrm{x}^{\mathrm{2}} −\mathrm{4qx}+\mathrm{2q}^{\mathrm{2}} −\mathrm{r}=\mathrm{0},\mathrm{roots}\:\mathrm{are}\:\mathrm{real}.\left(\mathrm{to}\:\mathrm{be}\:\mathrm{proved}\right) \\ $$$$−−−−−−−−−−− \\ $$$$\alpha+\beta=−\mathrm{p}\:,\alpha\beta=\mathrm{q} \\ $$$$\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} =\mathrm{r},\alpha^{\mathrm{2}} \beta^{\mathrm{2}} =\mathrm{s}=\mathrm{q}^{\mathrm{2}} \\ $$$$\mathrm{x}^{\mathrm{2}} −\mathrm{4qx}+\mathrm{2q}^{\mathrm{2}} −\mathrm{r}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\Rightarrow\mathrm{x}^{\mathrm{2}} −\mathrm{4}\left(\alpha\beta\right)\mathrm{x}+\mathrm{2}\left(\alpha\beta\right)^{\mathrm{2}} −\left(\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\:\:\mathrm{x}=\frac{\mathrm{4}\alpha\beta\pm\sqrt{\mathrm{16}\alpha^{\mathrm{2}} \beta^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}\right)\left(\mathrm{2}\left(\alpha\beta\right)^{\mathrm{2}} −\left(\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} \right)\right)}}{\mathrm{2}} \\ $$$$\:\:\mathrm{x}=\frac{\mathrm{4}\alpha\beta\pm\mathrm{2}\sqrt{\mathrm{4}\alpha^{\mathrm{2}} \beta^{\mathrm{2}} −\mathrm{2}\left(\alpha\beta\right)^{\mathrm{2}} +\left(\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} \right)}}{\mathrm{2}} \\ $$$$\:\:=\mathrm{2}\alpha\beta\pm\sqrt{\mathrm{4}\alpha^{\mathrm{2}} \beta^{\mathrm{2}} −\mathrm{2}\alpha^{\mathrm{2}} \beta^{\mathrm{2}} +\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} } \\ $$$$\:\:=\mathrm{2}\alpha\beta\pm\sqrt{\mathrm{2}\alpha^{\mathrm{2}} \beta^{\mathrm{2}} +\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} } \\ $$$$\alpha^{\mathrm{2}} ,\beta^{\mathrm{2}} \geqslant\mathrm{0}\Rightarrow\mathrm{2}\alpha^{\mathrm{2}} \beta^{\mathrm{2}} +\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$$\mathrm{Hence}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{are}\:\mathrm{real}. \\ $$
Commented by Tinkutara last updated on 19/Aug/17
Thank you.
$$\mathrm{Thank}\:\mathrm{you}. \\ $$

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