Question

Solve the inequality.

$log_3 \dfrac{|x^2-4x|+3}{x^2+|x-5|} \geq 0$

Answer 1

ANSWER:

Determine the define range of the question

And rewrite the inequality

We get

|x² - 4x| + 3/x² + |x - 5| ≥ 3⁰

Using , a⁰ = 1

→ |x² - 4x| + 3/x² + |x - 5| ≥ 1

→ (|x² - 4x| + 3/x² + |x - 5|) - 1 ≥ 0

→ |x² - 4x| + 3 - (x² + |x - 5|)/x² + |x - 5| ≥ 0

→ |x² - 4x| + 3 - x² - |x - 5|/x² + |x - 5| ≥ 0

→ |x² - 4x| + 3 - x² - |x - 5| ≥ 0

Seperating into possible cases

x² - 4x + 3 - x² - (x - 5)≥ 0 , x² - 4x ≥ 0 , x - 5 ≥ 0

- (x² - 4x) + 3 - x² - (x - 5)≥ 0 , x² - 4x < 0 , x - 5≥ 0

x² - 4x + 3 - x² - [ - (x - 5)]≥ 0 , x² - 4x≥ 0 , x - 5< 0

- (x² - 4x) + 3 - x² - [ - (x - 5)] ≥ 0 , x² - 4x < 0 ,x- 5<0

Solving the inequalities , we get

$\small{ \sf{x\leq \dfrac{8}{5} \: , \: x \in ( - \infty ,0 ] \cup}} [4 , + \infty ) , \sf{x \geq 5} \\ \\ \small{ \sf{ x \in [ \frac{3 - \sqrt{73} }{4},\frac{3 + \sqrt{73} }{ 4 } ],x \in(4,0) x \geq5}}$

$\small \sf{x \in - \dfrac{2}{3},x \in( - \infty,0 } ] \cup [4, + \infty ),x < 5$

$\small{ \sf{x \in[ \frac{1}{2}, 2] ,x \in(0,4),x < 5}}$

Find the intersection

x ≤ 8/5 , x belongs to [5 , + ∞]

x belongs to [3 - √73/4 , 3 + √73/4] , x belong to∅

x ≤ - 2/3 , x belongs to ( -∞ , 0] U [ 4 ,5)

x belongs to [½ , 2] , x belongs to (0 , 4)

Find the intersection

x belongs to ∅

x belongs to ∅

x belongs to ( - ∞ , - 2/3]

x belongs to [½ , 2]

Find the union

x belongs to ( - ∞ , - ⅔ ] U [½ , 2]

Anyways great & tough question

Answer 2

Given :   log₃  ( |x² - 4x |  + 3 )/(x² + | x - 5 | ) ≥ 0

To find : value of x

Solution:

log₃  ( |x² - 4x |  + 3 )/(x² + | x - 5 | ) ≥ 0

=> | x²  - 4x|  + 3   ≥  x²  + | x  - 5 |     as 3⁰ = 1

x²  - 4x = 0  x(x - 4) = 0

x = 0 , x = 4

x - 5 = 0  

So we get cases

x < 0    0 < x < 4  ,  4 < x < 5   & x  > 5

| x |  = x   if  x ≥ 0   &   -x  if x < 0

Case 1   x <  0

x²  - 4x  + 3 ≥   x²  +  5 - x

=> -3x  ≥  2

=>  x ≤ -2/3

Case 2  0 < x < 4

4x - x² + 3 ≥   x²  +  5 - x

=> 2x²  - 5x  + 2 ≤  0

=> 2x² - 4x  -x  + 2 ≤  0

=> (2x - 1)(x - 2 ) ≤ 0

1/2 ≤ x  ≤ 2  

Case 3   :  4 < x < 5

x²  - 4x  + 3 ≥   x²  +  5 - x  

already done above x ≤ -2/3

hence no solution for 4 < x < 5

case 4 : x > 5

x²  - 4x  + 3 ≥   x²  +  x  - 5

=>  -5x ≥  - 8

=> x ≤  8/5

hence no solution here

x ≤ -2/3   ,   1/2 ≤ x  ≤ 2  

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