Math, asked by Anonymous, 1 day ago

Solve -x2 + 3x – 2 ≥ 0

Question By ▄︻デ══━一 ‎ ‎ Nafees ⁍ ‎ ‎​

Answers

Answered by ItzImran
4

Answer:

Solution:

-x2 + 3x – 2 ≥ 0 ⇒ x2 – 3x + 2 ≤ 0

(x – 1) (x – 2) ≤ 0

[(x – 1) (x – 2) = 0

⇒ x = 1 or 2.

Here α = 1 and β = 2.

Note that α < β]

So for the inequality (x – 1) (x – 2) ≤ 2

x lies between 1 and 2

(i.e.) x ≥ 1 and x ≤ 2 or x ∈ [1, 2] or 1 ≤ x ≤ 2

Answered by mishrapiyushh1234
1

Answer:

First factor out the negative from the trinomial:

-(x^2–3x+2) >= 0

Now, factor the trinomial:

-(x-2)(x-1) >= 0

There are two solutions that make this a true statement: x = 2 or x = 1. The proper way of finding those solutions is to set one set of parentheses equal to zero. Then multiply by the other set of parentheses, and then multiply the negative and the result will be zero.

(x-2) = 0

Add 2 to both sides:

x = 1

x-1 = 0

Add 1 to both sides:

x = 1

Now substitute for each solution in the original inequality to prove this:

-x^2+3x-2>= 0

x = 1

-(1^2)+3*1–2>= 0

-1 + 3 - 2 >= 0

2 - 2 >= 0

0 >= 0

So that’s one valid solution! Now let’s check the other:

-x^2+3x-2>= 0

x = 2

-(2^2)+3*2–2 >= 0

-4+6–2 >= 0

2–2>= 0

0 >= 0

So x has at least two solutions: x >=1 Or x >= 2. Those are the two we solved for in the equation. However, the equation stipulates that the result can be greater than or equal to zero. Let’s substitute x for another value: 5.

-x^2+3x-2>= 0

x = 5

-(5^2)+3*5–2>= 0

-25+15–2>= 0

-10–2>= 0

-12>= 0

So 5 doesn’t work. How about 3?

-x^2+3x-2>= 0

-(3^2)+3*3–2>= 0

-9+9–2>= 0

0–2>= 0

-2>=0

So that doesn’t work either. How about if we set x itself equal to zero?

-x^2+3x-2>= 0

x = 0

0+0–2>= 0

-2>= 0

So that one doesn’t work either. So x = 0, x = 3, and x = 5 do not result in valid solutions, but we know that x = 1 and x = 2 work. How about any solution between 1 and 2? Let’s try 1.5:

-x^2+3x-2>= 0

x = 1.5

-(1.5^2)+3*1.5–2>= 0

-2.25+4.5–2>= 0

2.25–2>= 0

0.25>= 0

So x = 1.5 works.

So what does this mean for us? It means we have to update our original declaration of the solution. We know x = 1 and x = 2 are valid solutions, but we also know that any values for x less than 1 or greater than 2 are not valid.

Therefore, the solution is:

All real numbers, 1 =< x =< 2

Step-by-step explanation:

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