# Solve the inequality 2x^2-3x+1=<0

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The inequality `2x^2-3x+1=<0` has to be solved.

`2x^2-3x+1=<0`

`2x^2-2x-x+1=<0`

`2x(x - 1) - 1(x - 1)=<0`

`(2x - 1)(x - 1)=<0`

Now if the product has to be less than or equal to 0, either of the terms should be negative and the other should be positive

`2x - 1 >= 0` and` x - 1 <= 0`

`x >= 1/2` and `x <= 1`

`2x - 1 <= 0` and `x - 1 >= 0`

`x<= 1/2` and `x >= 1`

This is not possible as x cannot be less than 1/2 and greater than 1 at the same time.

The solution of the equation is `1/2 <= x <= 1`

According to the rule, between the values of the roots of the equation 2x^2-3x+1=0, the expression has the opposite sign to the sign of the leading coefficient. Outside the roots, the expresison has the same sign as the sign of the leading coefficient.

We'll identify the sign of the leading coefficient a = 2, so the sign is "+".

Therefore, the expression is negative between the value of the roots.

We'll detemrine the roots of the equation 2x^2-3x+1=0.

We'll apply the quadratic formula:

x1 = [3+sqrt(9-8)]/4

x1 = (3+1)/4

x1 = 1

x2 = (3-1)/4

x2 = 1/2

**The expression 2x^2-3x+1 is negative, 2x^2-3x+1 =< 0, if x belongs to the interval [1/2 ; 1].**