Question Number 78888 by zainal tanjung last updated on 21/Jan/20
$$\mathrm{The}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\left({x}−{a}\right)\left({x}−{b}\right)={abx}^{\mathrm{2}} \:\mathrm{are}\:\mathrm{always} \\ $$
Commented by mr W last updated on 21/Jan/20
$$...{are}\:{always}\:{what}? \\ $$
Answered by behi83417@gmail.com last updated on 21/Jan/20
![(1−ab)x^2 −(a+b)x+ab=0 △=(a+b)^2 −4ab(1−ab)= =a^2 +b^2 +2ab−4ab+4a^2 b^2 = =(a−b)^2 +4a^2 b^2 >0 [∀ a,b∈R] ⇒there is alweys two real roots.](Q78930.png)
$$\left(\mathrm{1}−\mathrm{ab}\right)\mathrm{x}^{\mathrm{2}} −\left(\mathrm{a}+\mathrm{b}\right)\mathrm{x}+\mathrm{ab}=\mathrm{0} \\ $$$$\bigtriangleup=\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{2}} −\mathrm{4ab}\left(\mathrm{1}−\mathrm{ab}\right)= \\ $$$$=\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{2ab}−\mathrm{4ab}+\mathrm{4a}^{\mathrm{2}} \mathrm{b}^{\mathrm{2}} = \\ $$$$=\left(\mathrm{a}−\mathrm{b}\right)^{\mathrm{2}} +\mathrm{4a}^{\mathrm{2}} \mathrm{b}^{\mathrm{2}} >\mathrm{0}\:\left[\forall\:\mathrm{a},\mathrm{b}\in\mathrm{R}\right] \\ $$$$\Rightarrow\mathrm{there}\:\mathrm{is}\:\mathrm{alweys}\:\mathrm{two}\:\mathrm{real}\:\mathrm{roots}. \\ $$