Solve for x : x^2 - 5x + 6 =< 0

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To solve the inequality above, first we have to calculate the roots of the equation x^2 - 5x + 6 = 0.

After finding the roots of the equation, we could write the expression in a factored form as:

1*(x-x1)(x-x2) =< 0

So, let's apply the quadratic formula to calculate the roots:

x1 = [5+sqrt(25-24)]/2

x1 = (5+1)/2

x1 = 6/2

**x1 = 3**

x2 = (5-1)/2

x2 = 4/2

**x2 = 2**

The inequality will be written as:

(x - 3)(x - 2) =< 0

Now, we'll discuss the inequality:

- the product is negative if one factor is positive and the other is negative:

x - 3>=0

We'll add 3 both sides:

x > =3

and

x - 2 =< 0

x =< 2

The common solution is the empty set.

Now, we'll consider the other alternative:

x - 3 =< 0

x =< 3

and

x - 2 >= 0

x >= 2

So, x belongs to the interval [2 , 3].

**Finally, the solution of the inequality is the inetrval identified above: ****[2 , 3].**

x^2-5x+6= 0. To solve for x.

Solution:

We can factorise left and equate each factor to zero and get x.

But try to solve by the method completing the square:

x^2-5x +6 = 0.

x^2 -5x +(5/2)^2 - (5/2)^2 +6 = 0, Added and subtracted (5/2)^2.

(x-5/2)^2- 25/4+6 = 0

(x-5/2)^2 =25/4 - 6 = (25-24)/4 = 1/4 = (1/2)^2.

(x-5/2)^2 = (1/2). Take square root:

x - 5/2 = 1/2 , or -1/2

x= 5/2+1/2 or 5/2 - 1/2

x = 3 or 2.

x^2 - 5x + 6 =< 0

First let us factor:

(x-3)(x-2) =< 0

Then we have two cases:

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)

then the solution is:

**x = [2, 3]**

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