Find the complete solution to the inequality :
x^3 – 8 ≤ 7x – 14
(A) x ≤ – 3 or x ≥ 1 (B) – 3 ≤ x ≤ 1
(C) – 3 ≤ x ≤ 1 or x ≥ 2 (D) x ≤ – 3 or 1 ≤ x ≤ 2
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Answered by
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x³ - 8 ≤ 7x - 14
x³ - 7x - 6 ≤ 0
let this equation = x³ - 7 x -6 = 0 then find the number of x whick makes it 0 then put x = 3 because if we put x = 3 we get 3³ - 7×3 - 6 = 0 rytee
then x = 3 x- 3 is a factor of this equation u already know because tumne isko 9th me pdha tha theek h na now u get it
divide this equation by x- 3
x³ - 7x - 6 ÷ x+3
do it ur self
after solving it we get (x-1)(x-2)(x+3) are the zeros of this equation
on representing these numbers on number line u get 4 cases first
∞ < x ≤ -3 second -3≤ x < 1 third 1 ≤ x <2 forth 2 ≤ x ∞
on checking it we will analyse that x ≤ - 3 then we can say that than we can say that it implies negative than value of x lies for (- ∞ to -3]
second case [-3 to 1 ] third x ≥ 2 means x lies for [2 to ∞ ] fourth x ≤ -3 means x lies from [- 3 to - ∞) we take only commen part in our answer
so commen part it (-∞ , -3] union [-3 , ∞)
x³ - 7x - 6 ≤ 0
let this equation = x³ - 7 x -6 = 0 then find the number of x whick makes it 0 then put x = 3 because if we put x = 3 we get 3³ - 7×3 - 6 = 0 rytee
then x = 3 x- 3 is a factor of this equation u already know because tumne isko 9th me pdha tha theek h na now u get it
divide this equation by x- 3
x³ - 7x - 6 ÷ x+3
do it ur self
after solving it we get (x-1)(x-2)(x+3) are the zeros of this equation
on representing these numbers on number line u get 4 cases first
∞ < x ≤ -3 second -3≤ x < 1 third 1 ≤ x <2 forth 2 ≤ x ∞
on checking it we will analyse that x ≤ - 3 then we can say that than we can say that it implies negative than value of x lies for (- ∞ to -3]
second case [-3 to 1 ] third x ≥ 2 means x lies for [2 to ∞ ] fourth x ≤ -3 means x lies from [- 3 to - ∞) we take only commen part in our answer
so commen part it (-∞ , -3] union [-3 , ∞)
Answered by
0
x³ - 8 ≤ 7x - 14
x³ - 7x - 6 ≤ 0
let this equation = x³ - 7 x -6 = 0 then find the number of x whick makes it 0 then put x = 3 because if we put x = 3 we get 3³ - 7×3 - 6 = 0 rytee
then x = 3 x- 3 is a factor of this equation u already know because tumne isko 9th me pdha tha theek h na now u get it
divide this equation by x- 3
x³ - 7x - 6 ÷ x+3
do it ur self
after solving it we get (x-1)(x-2)(x+3) are the zeros of this equation
on representing these numbers on number line u get 4 cases first
∞ < x ≤ -3 second -3≤ x < 1 third 1 ≤ x <2 forth 2 ≤ x ∞
on checking it we will analyse that x ≤ - 3 then we can say that than we can say that it implies negative than value of x lies for (- ∞ to -3]
second case [-3 to 1 ] third x ≥ 2 means x lies for [2 to ∞ ] fourth x ≤ -3 means x lies from [- 3 to - ∞) we take only commen part in our answer
so commen part it (-∞ , -3] union [-3 , ∞)
x³ - 7x - 6 ≤ 0
let this equation = x³ - 7 x -6 = 0 then find the number of x whick makes it 0 then put x = 3 because if we put x = 3 we get 3³ - 7×3 - 6 = 0 rytee
then x = 3 x- 3 is a factor of this equation u already know because tumne isko 9th me pdha tha theek h na now u get it
divide this equation by x- 3
x³ - 7x - 6 ÷ x+3
do it ur self
after solving it we get (x-1)(x-2)(x+3) are the zeros of this equation
on representing these numbers on number line u get 4 cases first
∞ < x ≤ -3 second -3≤ x < 1 third 1 ≤ x <2 forth 2 ≤ x ∞
on checking it we will analyse that x ≤ - 3 then we can say that than we can say that it implies negative than value of x lies for (- ∞ to -3]
second case [-3 to 1 ] third x ≥ 2 means x lies for [2 to ∞ ] fourth x ≤ -3 means x lies from [- 3 to - ∞) we take only commen part in our answer
so commen part it (-∞ , -3] union [-3 , ∞)
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