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Question Number 5638 by Daily last updated on 23/May/16
If a≤ 0, then the real values of x satisfying x^2 −2a ∣ x−a ∣−3a^2 =0 are
$$\mathrm{If}\:{a}\leqslant\:\mathrm{0},\:\mathrm{then}\:\mathrm{the}\:\mathrm{real}\:\mathrm{values}\:\mathrm{of}\:{x} \\ $$$$\mathrm{satisfying}\:{x}^{\mathrm{2}} −\mathrm{2}{a}\:\mid\:{x}−{a}\:\mid−\mathrm{3}{a}^{\mathrm{2}} =\mathrm{0}\:\mathrm{are} \\ $$
Answered by Yozzii last updated on 23/May/16
We can look for solutions based on intervals that cover the entire real line. a≤0 is some fixed real value. If x−a>0 or x>a⇒∣x−a∣=x−a. ∴ x^2 −2a(x−a)−3a^2 =0 x^2 −2ax−a^2 =0 x^2 −2ax+a^2 −2a^2 =0 (x−a)^2 =2a^2 ⇒x=a±(√2)(√a^2 ) x=a±(√2)∣a∣ which are real values of x. a≤0⇒∣a∣=−a⇒x=(1±(√2))a Now x>a⇒ x≠(1+(√2))a but x=(1−(√2))a since a≤0.{a≤0⇒(1−(√2))a≥0≥a or x≥a} If x=a≤0 the equation gives a^2 −3a^2 =0⇒2a^2 =0. This is true only if a=0. ∴ x=0 is possible. If x−a<0⇒x<a⇒∣x−a∣=a−x. ∴ x^2 −2a(a−x)−3a^2 =0 x^2 +2ax−5a^2 =0 x^2 +2ax+a^2 −6a^2 =0 (x+a)^2 =6a^2 ⇒x=−a±(√6)∣a∣=a(−1±(√6)) Since a≤0 and x<a⇒x≠(−1−(√6))a but x=a((√6)−1). ∴ If a=0 the quadratic equation has two equal roots then x=0. If a≠0 then x=((√6)−1)a or x=(1−(√2))a.
$${We}\:{can}\:{look}\:{for}\:{solutions}\:{based}\:{on} \\ $$$${intervals}\:{that}\:{cover}\:{the}\:{entire}\:{real}\:{line}. \\ $$$${a}\leqslant\mathrm{0}\:{is}\:{some}\:{fixed}\:{real}\:{value}. \\ $$$${If}\:{x}−{a}>\mathrm{0}\:{or}\:{x}>{a}\Rightarrow\mid{x}−{a}\mid={x}−{a}. \\ $$$$\therefore\:{x}^{\mathrm{2}} −\mathrm{2}{a}\left({x}−{a}\right)−\mathrm{3}{a}^{\mathrm{2}} =\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{ax}−{a}^{\mathrm{2}} =\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{ax}+{a}^{\mathrm{2}} −\mathrm{2}{a}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({x}−{a}\right)^{\mathrm{2}} =\mathrm{2}{a}^{\mathrm{2}} \\ $$$$\Rightarrow{x}={a}\pm\sqrt{\mathrm{2}}\sqrt{{a}^{\mathrm{2}} } \\ $$$${x}={a}\pm\sqrt{\mathrm{2}}\mid{a}\mid\:{which}\:{are}\:{real}\:{values}\:{of}\:{x}. \\ $$$${a}\leqslant\mathrm{0}\Rightarrow\mid{a}\mid=−{a}\Rightarrow{x}=\left(\mathrm{1}\pm\sqrt{\mathrm{2}}\right){a} \\ $$$${Now}\:{x}>{a}\Rightarrow\:{x}\neq\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){a}\:{but}\:{x}=\left(\mathrm{1}−\sqrt{\mathrm{2}}\right){a} \\ $$$${since}\:{a}\leqslant\mathrm{0}.\left\{{a}\leqslant\mathrm{0}\Rightarrow\left(\mathrm{1}−\sqrt{\mathrm{2}}\right){a}\geqslant\mathrm{0}\geqslant{a}\:{or}\:{x}\geqslant{a}\right\} \\ $$$$ \\ $$$${If}\:{x}={a}\leqslant\mathrm{0}\:{the}\:{equation}\:{gives} \\ $$$${a}^{\mathrm{2}} −\mathrm{3}{a}^{\mathrm{2}} =\mathrm{0}\Rightarrow\mathrm{2}{a}^{\mathrm{2}} =\mathrm{0}.\:{This}\:{is}\:{true}\:{only}\:{if}\:{a}=\mathrm{0}. \\ $$$$\therefore\:{x}=\mathrm{0}\:{is}\:{possible}. \\ $$$$ \\ $$$${If}\:{x}−{a}<\mathrm{0}\Rightarrow{x}<{a}\Rightarrow\mid{x}−{a}\mid={a}−{x}. \\ $$$$\therefore\:{x}^{\mathrm{2}} −\mathrm{2}{a}\left({a}−{x}\right)−\mathrm{3}{a}^{\mathrm{2}} =\mathrm{0} \\ $$$${x}^{\mathrm{2}} +\mathrm{2}{ax}−\mathrm{5}{a}^{\mathrm{2}} =\mathrm{0} \\ $$$${x}^{\mathrm{2}} +\mathrm{2}{ax}+{a}^{\mathrm{2}} −\mathrm{6}{a}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({x}+{a}\right)^{\mathrm{2}} =\mathrm{6}{a}^{\mathrm{2}} \\ $$$$\Rightarrow{x}=−{a}\pm\sqrt{\mathrm{6}}\mid{a}\mid={a}\left(−\mathrm{1}\pm\sqrt{\mathrm{6}}\right) \\ $$$${Since}\:{a}\leqslant\mathrm{0}\:{and}\:{x}<{a}\Rightarrow{x}\neq\left(−\mathrm{1}−\sqrt{\mathrm{6}}\right){a} \\ $$$${but}\:{x}={a}\left(\sqrt{\mathrm{6}}−\mathrm{1}\right). \\ $$$$ \\ $$$$\therefore\:{If}\:{a}=\mathrm{0}\:{the}\:{quadratic}\:{equation}\:{has} \\ $$$${two}\:{equal}\:{roots}\:{then}\:{x}=\mathrm{0}.\:{If}\:{a}\neq\mathrm{0}\:{then}\: \\ $$$${x}=\left(\sqrt{\mathrm{6}}−\mathrm{1}\right){a}\:{or}\:{x}=\left(\mathrm{1}−\sqrt{\mathrm{2}}\right){a}. \\ $$

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