Need help solving x^2-3x+4<0 and how do you use graph utilities to solve an equation?

Question

what is the basic method to solve inequalities?

Answer 1

Do this:

x^2-3*x+4=
=x^2-2*(3/2)*x+(9/4)+(7/4)=
=(x-3/2)^2+7/4

(This method is universal, you can make it on your own for ax^2+bx+c)


Obviously, (x-3/2)^2<-7/4 has no solution in the set of the real numbers. This method is good for graphing the parabola: it explicitely shows where is the extremum. In this case it is
T(3/2, 7/4). Since a>0 (from ax^2+bx+c), T is minimum, otherwise it would be maximum. So it should be obvious by now that there are no intersections with the x-axis, and therefore, no real solution. (The real solutions of ax^2+bx+c=0 can be read from the graph - the intersections of the parabola with the x-axis)
For inequalities, just look for the intervals that match (ON THE x-AXIS) for the part of the parabola below the x-axis (in this case) or above the x-axis, for inequalities ax^2+bx+c>0.

Answer 2

Inthis case, you have a parabola that opens 'upwards' (since the x² term is positive) so the only places whare the function is negative (<0) is between the roots (where
y = 0). The problem with this one is that the quadratic has no real roots and so there is no real value for x for which the function is ever less than 0. The minimum value the function ever achieves is 7/4 at x = 3/2.

Therefore there is *no* real-valued x for which the function is less than 0


Doug

Answer 3

x^2-3x+4<0
Since it cannot be factored
Put
x^2-3x+4=0
Solve using the quadratic formula
You will see that it does not have real roots
The imaginary roots are
{3+Sq-rt(-7)}/2 and {3 - Sqrt(-7)}/2