If a≤ 0, then the real values of x satisfying x^2 −2a ∣ x−a ∣−3a^2 =0 are


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Question Number 5638 by Daily last updated on 23/May/16

$If a ⩽ 0, then the real values of x$ $satisfying x^{2} - 2 a ∣ x - a ∣ - 3 a^{2} = 0 are$
	Answered by Yozzii last updated on 23/May/16
	$We can look for solutions based on intervals that cover the entire real line. a≤0 is some fixed real value. If x−a>0 or x>a⇒∣x−a∣=x−a. ∴ x^2 −2a(x−a)−3a^2 =0 x^2 −2ax−a^2 =0 x^2 −2ax+a^2 −2a^2 =0 (x−a)^2 =2a^2 ⇒x=a±(√2)(√a^2 ) x=a±(√2)∣a∣ which are real values of x. a≤0⇒∣a∣=−a⇒x=(1±(√2))a Now x>a⇒ x≠(1+(√2))a but x=(1−(√2))a since a≤0.{a≤0⇒(1−(√2))a≥0≥a or x≥a} If x=a≤0 the equation gives a^2 −3a^2 =0⇒2a^2 =0. This is true only if a=0. ∴ x=0 is possible. If x−a<0⇒x<a⇒∣x−a∣=a−x. ∴ x^2 −2a(a−x)−3a^2 =0 x^2 +2ax−5a^2 =0 x^2 +2ax+a^2 −6a^2 =0 (x+a)^2 =6a^2 ⇒x=−a±(√6)∣a∣=a(−1±(√6)) Since a≤0 and x<a⇒x≠(−1−(√6))a but x=a((√6)−1). ∴ If a=0 the quadratic equation has two equal roots then x=0. If a≠0 then x=((√6)−1)a or x=(1−(√2))a.$
	$W e c a n l o o k f o r s o l u t i o n s b a s e d o n$ $i n t e r v a l s t h a t c o v e r t h e e n t i r e r e a l l i n e .$ $a ⩽ 0 i s s o m e f i x e d r e a l v a l u e .$ $I f x - a > 0 o r x > a \Rightarrow∣ x - a ∣= x - a .$ $∴ x^{2} - 2 a (x - a) - 3 a^{2} = 0$ $x^{2} - 2 a x - a^{2} = 0$ $x^{2} - 2 a x + a^{2} - 2 a^{2} = 0$ ${(x - a)}^{2} = 2 a^{2}$ $\Rightarrow x = a \pm \sqrt{2} \sqrt{a^{2}}$ $x = a \pm \sqrt{2} ∣ a ∣ w h i c h a r e r e a l v a l u e s o f x .$ $a ⩽ 0 \Rightarrow∣ a ∣= - a \Rightarrow x = (1 \pm \sqrt{2}) a$ $N o w x > a \Rightarrow x \neq (1 + \sqrt{2}) a b u t x = (1 - \sqrt{2}) a$ $s i n c e a ⩽ 0 . {a ⩽ 0 \Rightarrow (1 - \sqrt{2}) a ⩾ 0 ⩾ a o r x ⩾ a}$ $I f x = a ⩽ 0 t h e e q u a t i o n g i v e s$ $a^{2} - 3 a^{2} = 0 \Rightarrow 2 a^{2} = 0 . T h i s i s t r u e o n l y i f a = 0 .$ $∴ x = 0 i s p o s s i b l e .$ $I f x - a < 0 \Rightarrow x < a \Rightarrow∣ x - a ∣= a - x .$ $∴ x^{2} - 2 a (a - x) - 3 a^{2} = 0$ $x^{2} + 2 a x - 5 a^{2} = 0$ $x^{2} + 2 a x + a^{2} - 6 a^{2} = 0$ ${(x + a)}^{2} = 6 a^{2}$ $\Rightarrow x = - a \pm \sqrt{6} ∣ a ∣= a (- 1 \pm \sqrt{6})$ $S i n c e a ⩽ 0 a n d x < a \Rightarrow x \neq (- 1 - \sqrt{6}) a$ $b u t x = a (\sqrt{6} - 1) .$ $∴ I f a = 0 t h e q u a d r a t i c e q u a t i o n h a s$ $t w o e q u a l r o o t s t h e n x = 0 . I f a \neq 0 t h e n$ $x = (\sqrt{6} - 1) a o r x = (1 - \sqrt{2}) a .$
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