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Question Number 59499 by MathGuru last updated on 11/May/19

Commented by maxmathsup by imad last updated on 11/May/19
$we have x^2 −3x =x(x−3) and 2x−6 =2(x−3) and x^2 −5x +6 =x^2 −9 −5x +15 =(x−3)(x+3) −5(x−3)=(x−3)(x−2) so (e) ⇔∣x∣∣x−3∣+2∣x−3∣ =∣x−3∣∣x−2∣ ⇔∣x−3∣( ∣x∣ +2−∣x−2∣)=0 ⇔ x=3 or ∣x∣−∣x−2∣ +2 =0 x −∞ 0 2 +∞ ∣x∣ −x 0 x x ∣x−2∣ −x+2 −x+2 x−2 f(x) 0 2x 4 if x≤0 all x is solution if 0≤x≤2 f(x)=0 ⇔2x=0 ⇔x=0 if x≥2 f(x)=0 ⇔4=0 impossible so the set of solution is]−∞,0]∪{3}$
$w e h a v e x^{2} - 3 x = x (x - 3) a n d 2 x - 6 = 2 (x - 3) a n d$ $x^{2} - 5 x + 6 = x^{2} - 9 - 5 x + 15 = (x - 3) (x + 3) - 5 (x - 3) = (x - 3) (x - 2) s o$ $(e) \Leftrightarrow∣ x ∣∣ x - 3 ∣ + 2 ∣ x - 3 ∣ =∣ x - 3 ∣∣ x - 2 ∣ \Leftrightarrow∣ x - 3 ∣ (∣ x ∣ + 2 - ∣ x - 2 ∣) = 0 \Leftrightarrow$ $x = 3 o r ∣ x ∣ - ∣ x - 2 ∣ + 2 = 0$ $x - \infty 0 2 + \infty$ $∣ x ∣ - x 0 x x$ $∣ x - 2 ∣ - x + 2 - x + 2 x - 2$ $f (x) 0 2 x 4$ $i f x ⩽ 0 a l l x i s s o l u t i o n$ $i f 0 ⩽ x ⩽ 2 f (x) = 0 \Leftrightarrow 2 x = 0 \Leftrightarrow x = 0$ $i f x ⩾ 2 f (x) = 0 \Leftrightarrow 4 = 0 i m p o s s i b l e s o t h e s e t o f s o l u t i o n i s] - \infty, 0] \cup {3}$
Commented by maxmathsup by imad last updated on 11/May/19

$f (x) =∣ x ∣ - ∣ x - 2 ∣ + 2$
Commented by MathGuru last updated on 11/May/19

$We may solve in a different mehod :$ $∣ x^{2} - 3 x ∣ + ∣ 2 x - 6 ∣ = ∣ x^{2} - 5 x + 6 ∣$ $\Rightarrow∣ x^{2} - 3 x ∣ + ∣ 6 - 2 x ∣ = ∣ (x^{2} - 3 x) + (6 - 2 x) ∣$ $\Rightarrow (x^{2} - 3 x) (6 - 2 x) ⩾ 0$ $\Rightarrow x (x - 3) 2 (3 - x) ⩾ 0$ $\Rightarrow 2 x {(x - 3)}^{2} ⩽ 0$ $\Rightarrow x ⩽ 0 or x = 3$
	Answered by tanmay last updated on 11/May/19

	$∣ x (x - 3) ∣ + ∣ 2 (x - 3) ∣=∣ (x - 2) (x - 3) ∣$ $c r i t i c a l v a l u e o f x = 0, 2, 3$ $w h e n x ⩾ 3$ $x (x - 3) + 2 (x - 3) = (x - 2) (x - 3)$ $x^{2} - 3 x + 2 x - 6 = x^{2} - 5 x + 6$ $4 x = 12 x = 3 . . . . l o o k h e r e$ $w h e n x < 0$ $x (x - 3) - 2 (x - 3) = (x - 2) (x - 3)$ $x^{2} - 3 x - 2 x + 6 = x^{2} - 5 x + 6$ $i t b e c o m e i d e n t i t y . . . . l o o k h e r e$ $w h e n 3 > x > 2$ $- x (x - 3) - 2 (x - 3) = - (x - 2) (x - 3)$ $- x^{2} + 3 x - 2 x + 6 = - x^{2} + 5 x - 6$ $- 4 x = - 12$ $x = 3 b u t t h i s v a l u e d o n o t c o n f o r m t o 3 > x > 2$ $w h e n 2 > x > 0$ $- x (x - 3) - 2 (x - 3) = (x - 2) (x - 3)$ $- x^{2} + 3 x - 2 x + 6 = x^{2} - 5 x + 6$ $- 2 x^{2} + 6 x = 0$ $- 2 x (x - 3) = 0$ $x = 0 a n d x = 3$ $b u t x = 0 a n d x = 3 d o n o t c o n f o r m t o$ $c o n d i t i o n 2 > x > 0$ $s o x = 3 i s t h e o n l y s o l u t i o n$ $p l s c h e c k$
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