# math

posted by
**please i need help.Dan** on
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x^2-3x+2<0

Put in y=x^2-3x+2 in your graphing calculator, and graph is. See where y is negative. That area is the solution

The graph will be a second degree equation, and probably cross the y=0 axis at the x defined in (x^2-3x+2=0)

Factor that equation and solve for those values of x. There will be two of them, and that will divide the x axis into three regions (to the left of the most negative, between the two values of x, and to the right of the highest value)

Test each region by picking an x and seeing if the value of y=x^2-3x+2 is negative.

(x-1)(x-2)

x=1 x=2

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