How do I deal with quadratic inequalities?

First of all, if not already done, you would arrange all the x's or whichever letter used on one side of the inequality. E.g. for x2+ 19x - 5 < 6x - 45 you would arrange it like: x2+ 13x + 40 < 0. Whilst you are still understanding these problems, I suggest you sketch a graph to help you. Therefore, for the moment, instead of an inequality, we can replace this with x2+13x+40=y. This is in the format of a line. So after factorising to get (x+8)(x+5)=y we can sketch the graph, and label in particular the places the line cuts the x axis, which are the solutions to the equation, x=-8 and x=-5. So now there's a graph and it's time to consider the inequality. The inequality is asking us to find the values of x, which, if plugged into the equation, result in a value less than zero. The line shows us exactly this very clearly. The values where(x+8)(x+5) is less than zero must be below the line y=0 (or the x axis). This is the part of the curve between x=-8 and x=-5. Hence the solution to the inequality is -8<x<-5. If the inequality is reversed, (x+8)(x+5)>0, we look for the part of the line above the x axis. This would be when x<-8 and when x>-5. It is important we write both these inequalities to get full marks.

Answered by Asha C. Maths tutor

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