Please answer the sum :-
x² - 5x + 6 ≤ 0 ; then , x ∈ (?)
So,
Find the value of (?).
© Note that (?) can contain multiple values and should be given in proper brackets.
Class 11 | Algebraic Inequalities.
NO SPAMS ALLOWED.
Answers
Answered by
2
Hola there,
=> x² - 5x + 6 ≤ 0
=> x² - 3x - 2x + 6 ≤ 0
=> x(x - 3) - 2(x - 3) ≤ 0
=> (x - 3)(x - 2) ≤ 0
Case 1:
(x - 3) ≤ 0 and (x - 2) ≥ 0
x ≤ 3 and x ≥ 2
=> x ∈ (2, 3)
Case 2:
(x - 3) ≥ 0 and (x - 2) ≤ 0
x ≥ 3 and x ≤ 2. ... Impossible sol.
So the ans will be ;
x ∈ [2, 3]
=> x² - 5x + 6 ≤ 0
=> x² - 3x - 2x + 6 ≤ 0
=> x(x - 3) - 2(x - 3) ≤ 0
=> (x - 3)(x - 2) ≤ 0
Case 1:
(x - 3) ≤ 0 and (x - 2) ≥ 0
x ≤ 3 and x ≥ 2
=> x ∈ (2, 3)
Case 2:
(x - 3) ≥ 0 and (x - 2) ≤ 0
x ≥ 3 and x ≤ 2. ... Impossible sol.
So the ans will be ;
x ∈ [2, 3]
HridayAg0102:
it is still wrong
ok then delete my answer.
u just need to put x ∈ [2,3]
sorry I can't edit it...
lemme give u option
Done!!
yes
now it is correct
sorry for the earlier mistake.
no problem
Answered by
4
Given Equation is x^2 - 5x + 6 ≤ 0.
(x - 2)(x - 3) ≤ 0
(x - 3) ≤ 0 and (x - 2) ≥ 0
x ≤ 3 and x ≥ 2
x ∈ [2,3].
Note:
Here,
x value lies between 2 and 3.
Therefore 2 ≤ x ≤ 3.
x ≤ 2 or x ≥ 3
Hope this helps!
(x - 2)(x - 3) ≤ 0
(x - 3) ≤ 0 and (x - 2) ≥ 0
x ≤ 3 and x ≥ 2
x ∈ [2,3].
Note:
Here,
x value lies between 2 and 3.
Therefore 2 ≤ x ≤ 3.
x ≤ 2 or x ≥ 3
Hope this helps!
correct
byt the above one has used wrong sequence of brackets
Which step?
last one
x ∈ [2,3] ? Is it incorrect?
yep
I was saying about the other answer
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