find the solution set of equation ∣x^2 −5x+4∣ = x^2 −5∣x∣ + 4


Question and Answers Forum

All Questions Topic List
Coordinate Geometry Questions
Previous in All Question Next in All Question
Previous in Coordinate Geometry Next in Coordinate Geometry
Question Number 106356 by bobhans last updated on 04/Aug/20

$find the solution set of equation$ $∣ x^{2} - 5 x + 4 ∣ = x^{2} - 5 ∣ x ∣ + 4$
Commented by bobhans last updated on 05/Aug/20

$thanm you both$
	Answered by 1549442205PVT last updated on 05/Aug/20
	$Set f(x)=x^2 −5x+4=(x−1)(x−4) i)If 1≤x≤4 then f(x)≤0 and ∣x∣=x⇒ ∣x^2 −5x+4∣ = x^2 −5∣x∣ + 4 (1) ⇔−(x^2 −5x+4)=x^2 −5x+4 ⇔x^2 −5x+4=0⇔(x−1)(x−4)=0 ⇔x∈{1;4} ii)If x>4 then f(x)>0. (1)⇔x^2 −5x+4= x^2 −5x + 4 ⇔0.x=0⇒x∈(4;+∞) iii)If 0≤x<1 then f(x)>0,∣x∣=x (1)⇔x^2 −5x+4 = x^2 −5x + 4 ⇔0.x=0⇒x∈[0;1) iv)If −∞<x<0⇒f(x)>0,∣x∣=−x (1)⇔x^2 −5x+4=x^2 +5x+4=0 ⇔10x=0⇔x=0⇒has no roots Thus,solution set of the given equation is S=[0,1]∪[4;+∞)$
	$Set f (x) = x^{2} - 5 x + 4 = (x - 1) (x - 4)$ $i) If 1 ⩽ x ⩽ 4 then f (x) ⩽ 0 and ∣ x ∣= x \Rightarrow$ $∣ x^{2} - 5 x + 4 ∣ = x^{2} - 5 ∣ x ∣ + 4 (1)$ $\Leftrightarrow - (x^{2} - 5 x + 4) = x^{2} - 5 x + 4$ $\Leftrightarrow x^{2} - 5 x + 4 = 0 \Leftrightarrow (x - 1) (x - 4) = 0$ $\Leftrightarrow x \in {1; 4}$ $ii) If x > 4 then f (x) > 0 .$ $(1) \Leftrightarrow x^{2} - 5 x + 4 = x^{2} - 5 x + 4$ $\Leftrightarrow 0 . x = 0 \Rightarrow x \in (4; + \infty)$ $iii) If 0 ⩽ x < 1 then f (x) > 0, ∣ x ∣= x$ $(1) \Leftrightarrow x^{2} - 5 x + 4 = x^{2} - 5 x + 4$ $\Leftrightarrow 0 . x = 0 \Rightarrow x \in [0; 1)$ $iv) If - \infty < x < 0 \Rightarrow f (x) > 0, ∣ x ∣= - x$ $(1) \Leftrightarrow x^{2} - 5 x + 4 = x^{2} + 5 x + 4 = 0$ $\Leftrightarrow 10 x = 0 \Leftrightarrow x = 0 \Rightarrow has no roots$ $Thus, solution set of the given equation$ $is S = [0, 1] \cup [4; + \infty)$
	Answered by mathmax by abdo last updated on 04/Aug/20
	$e⇒∣x^2 −5x+4∣−x^2 +5∣x∣−4 =0 =A(x) x^2 −5x+4 =x^2 −4x−(x−4) =x(x−4)−(x−4)=(x−4)(x−1) x −∞ 0 1 4 +∞ ∣x^2 −5x+4∣ x^2 −5x+4 −x^2 +5x−4 −x^2 +5x−4 x^2 −5x+4 ∣x∣ −x 0 x x x A(x) −10x −2x^2 +10x−8 s.v 0 so A(x)= { ((−10x if x≤0)),((−2x^2 +10x−8 if 0≤x≤4)) :} and A(x) =0 if x≥4 x≤0 ⇒−10x =0 ⇒x=0 0≤x≤4 ⇒−2x^2 +10x −8 =0 ⇒2x^2 −10x +8 =0 ⇒ x^2 −5x +4 =0 ⇒x =1 or x=4 ⇒ set of solution is S ={0,1} ∪[4,+∞[$
	$e \Rightarrow∣ x^{2} - 5 x + 4 ∣ - x^{2} + 5 ∣ x ∣ - 4 = 0 = A (x)$ $x^{2} - 5 x + 4 = x^{2} - 4 x - (x - 4) = x (x - 4) - (x - 4) = (x - 4) (x - 1)$ $x - \infty 0 1 4 + \infty$ $∣ x^{2} - 5 x + 4 ∣ x^{2} - 5 x + 4 - x^{2} + 5 x - 4 - x^{2} + 5 x - 4 x^{2} - 5 x + 4$ $∣ x ∣ - x 0 x x x$ $A (x) - 10 x - {2 x}^{2} + 10 x - 8 s . v 0$ $so A (x) = {\begin{cases} - 10 x if x ⩽ 0 \\ - {2 x}^{2} + 10 x - 8 if 0 ⩽ x ⩽ 4 \end{cases}$ $and A (x) = 0 if x ⩾ 4$ $x ⩽ 0 \Rightarrow - 10 x = 0 \Rightarrow x = 0$ $0 ⩽ x ⩽ 4 \Rightarrow - {2 x}^{2} + 10 x - 8 = 0 \Rightarrow {2 x}^{2} - 10 x + 8 = 0 \Rightarrow$ $x^{2} - 5 x + 4 = 0 \Rightarrow x = 1 or x = 4 \Rightarrow$ $set of solution is S = {0, 1} \cup [4, + \infty [$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum