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Alinara [238K]
3 years ago
10

F(x) = -1/2x^2 + 3/2x - 2 I need to find the zeros​

Mathematics
1 answer:
Pachacha [2.7K]3 years ago
4 0

Answer:

{-1, 4}

Step-by-step explanation:

In the given f(x) = -1/2x^2 + 3/2x - 2

set f(x) = 0 and then multiply all the numeric terms by -2 to remove the fractions:

f(x) = -1/2x^2 + 3/2x - 2 = 0  =>  1x^2 - 3x - 4 = 0  

This factors into (x - 4)(x + 1) = 0, whose roots are {-1, 4}.  These are the zeros.

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Answer:

(b)  (x + 1 ≤ 1) ∩ (x + 1 ≥ 1) =  {x | x = 0}

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Here, the given expression is : {x | x = 0}

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Adding -1 BOTH sides, we get:

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So, (x + 1 < -1) ∩ (x + 1 < 1) =  { -∞ , .... , -4,-3}∩ { -∞ , .... , -4,-3,-2,-1}  

=  { -∞ , .... , -4,-3}

⇒ (x + 1 < -1) ∩ (x + 1 < 1) ≠ {0}

Similarly, solving

(b) (x + 1 ≤ 1) ∩ (x + 1 ≥ 1)

(x + 1 ≤ 1) =  x≤ 0 =   { -∞ , .... , -4,-3,-2,-1, 0}

(x + 1 ≥ 1) =  x ≥ 0 =  {0,1,2,3,... ∞}

⇒ (x + 1 ≤ 1) ∩ (x + 1 ≥ 1) = {0}

(b)(x + 1 < 1) ∩ (x + 1 > 1)

(x + 1 < 1) =  x <  0 =   { -∞ , .... , -4,-3,-2,-1}

(x + 1 > 1) =  x  > 0 =  { 1,2,3,... ∞}

⇒ (x + 1 ≤ 1) ∩ (x + 1 ≥ 1) = Ф

Hence,  (x + 1 ≤ 1) ∩ (x + 1 ≥ 1) =  {x | x = 0}

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