find-the-solution-set-of-equation-x-2-5x-4-x-2-5-x-4-

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} - 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 \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 [