Quote:
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).
Yes, this solves my doubt. You're suggesting that whatever the equation is convert it into the form ax^2+bx+c such that the "a" (coeff of x^2) is always +ve. So, for example, if we have to find where is the inverted-V curve +ve, we are actually finding where is the V-curve(mirror image) negative.
My only question is: Can we safely assume that a curve with a +ve coeff of x^2 is always a V-shaped curve?