How to solve quadratic inequalities - Graphic approach.\(x^2-4x+3<0\) is the graph of parabola and it look likes this:
Attachment:
en.plot (1).png
Intersection points are the roots of the equation \(x^2-4x+3=0\), which are \(x_1=1\) and \(x_2=3\).
"<" sign means in which range of \(x\) the graph is
below x-axis. Answer is \(1<x<3\) (between the roots).
If the sign were ">": \(x^2-4x+3>0\). First find the roots (\(x_1=1\) and \(x_2=3\)).
">" sign means in which range of \(x\) the graph is
above x-axis. Answer is \(x<1\) and \(x>3\) (to the left of the smaller root and to the right of the bigger root).
This approach works for any quadratic inequality. For example: \(-x^2-x+12>0\), first rewrite this as \(x^2+x-12<0\) (so that the coefficient of x^2 to be positive. It's possible to solve without rewriting, but easier to master one specific pattern).
\(x^2+x-12<0\). Roots are \(x_1=-4\) and \(x_1=3\) --> below (
"<") the x-axis is the range for \(-4<x<3\) (between the roots).
Again if it were \(x^2+x-12>0\), then the answer would be \(x<-4\) and \(x>3\) (to the left of the smaller root and to the right of the bigger root).
Hope it helps.