Question Number 70874 by Mr. K last updated on 09/Oct/19
$${If}\:\mathrm{4}−\mathrm{2}\sqrt{\mathrm{5}}\:{and}\:\mathrm{4}+\mathrm{2}\sqrt{\mathrm{5}\:}\:{are}\:{solutions} \\ $$$${of}\:{x}^{\mathrm{2}} +\left(\mathrm{5}{a}−{b}\right){x}+\left(\mathrm{3}{b}−{a}\right)=\mathrm{0} \\ $$$${whete}\:{a}\:{and}\:{b}\:{are}\:{real}\:{numbers},\: \\ $$$${determine}\:{the}\:{product}\:{of}\:\boldsymbol{{ab}}. \\ $$
Answered by tw000001 last updated on 09/Oct/19
$${x}=\mathrm{4}\pm\mathrm{2}\sqrt{\mathrm{5}}\rightarrow\left({x}−\mathrm{4}\right)^{\mathrm{2}} =\mathrm{20} \\ $$$$\rightarrow{x}^{\mathrm{2}} −\mathrm{8}{x}−\mathrm{4}=\mathrm{0} \\ $$$$\rightarrow\begin{cases}{\mathrm{5}{a}−{b}=−\mathrm{8}}\\{{a}−\mathrm{3}{b}=\mathrm{4}}\end{cases} \\ $$$$\rightarrow\left({a},{b}\right)=\left(−\mathrm{2},−\mathrm{2}\right) \\ $$$$\therefore{ab}=\mathrm{4} \\ $$
Answered by Rasheed.Sindhi last updated on 09/Oct/19
$$\mathbb{A}{n}\mathbb{O}{ther}\mathbb{W}{ay} \\ $$$${Say}\:\alpha,\beta\:{are}\:{roots}. \\ $$$$\alpha+\beta=\mathrm{5}{a}−{b}=\left(\mathrm{4}−\mathrm{2}\sqrt{\mathrm{5}}\:\right)+\:\left(\mathrm{4}+\mathrm{2}\sqrt{\mathrm{5}\:}\:\right)=\mathrm{8} \\ $$$$\alpha\beta=\mathrm{3}{b}−{a}=\left(\mathrm{4}−\mathrm{2}\sqrt{\mathrm{5}}\:\right)×\:\left(\mathrm{4}+\mathrm{2}\sqrt{\mathrm{5}\:}\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{4}\right)^{\mathrm{2}} −\left(\mathrm{2}\sqrt{\mathrm{5}}\:\right)^{\mathrm{2}} =\mathrm{16}−\mathrm{20}=−\mathrm{4} \\ $$$$\mathrm{5}{a}−{b}=\mathrm{8}\:\wedge\:\mathrm{3}{b}−{a}=−\mathrm{4} \\ $$$$\left({a},{b}\right)=\left(−\mathrm{2},−\mathrm{2}\right) \\ $$$${ab}=\left(−\mathrm{2}\right)\left(−\mathrm{2}\right)=\mathrm{4} \\ $$