 Lesson

Let's say we have the inequality $x+4<2x-6$x+4<2x6.

By now we know that we can use our elementary operations $+$+,$-$,$\times$×,$\div$÷​ to solve inequalities like this in exactly the same way that we solve equations, as long as we remember to reverse the inequality sign whenever we multiply or divide by a negative number. Hence, we could get our answer like this:

 $x+4$x+4 $<$< $2x-6$2x−6 $x$x $<$< $2x-10$2x−10 $-x$−x $<$< $-10$−10 $x$x $>$> $10$10

But what question are we really being asked when we are asked to solve $x+4<2x-6$x+4<2x6? We are being asked for values of $x$x at which $x+4$x+4 is less than $2x-6$2x6.

We can answer this question graphically by considering two lines $y_1=x+4$y1=x+4 and $y_2=2x-6$y2=2x6 Solving $x+4<2x-6$x+4<2x6 is the same as solving $y_1y1<y2. In other words, for what values of$x$x is the line$y_1$y1 below the line$y_2$y2? Sure enough, our graph confirms what we already knew: that$x+4<2x-6$x+4<2x6 for$x>10$x>10. ### Solving using graphs Now, what if we are given an inequality involving a quadratic, such as$x^2+2x-8>0$x2+2x8>0? Our elementary operations alone will be insufficient to solve this algebraically. We will need to use the above method to find our solutions using a graph. Let's use what we already know about quadratics to graph$y=x^2+2x-8$y=x2+2x8. We know that the parabola will be concave up, since the coefficient of$x^2$x2 is positive. We also get a$y$y-intercept of$y=-8$y=8 after substituting$x=0$x=0. We can use our vertex formula$x=\frac{-b}{2a}$x=b2a to find the axis of symmetry where our turning point lies, which we know will be a minimum turning point due to the concavity. $x$x$=$=$\frac{-b}{2a}$−b2a​$=$=$\frac{-2}{2\times1}$−22×1​$=$=$-1$−1 Last of all, we can also factorise the quadratic to find our$x$x-intercepts. $x^2+2x-8$x2+2x−8$=$=$0$0$\left(x+4\right)\left(x-2\right)$(x+4)(x−2)$=$=$0$0$x$x$=$=$-4$−4,$2$2 Now we can graph our parabola. Being asked to find solutions to$x^2+2x-8>0$x2+2x8>0 is exactly the same as being asked to find values of$x$x at which the graph of$y=x^2+2x-8$y=x2+2x8 is above the line$y=0$y=0 (the$x$x-axis). In other words, when is$y=x^2+2x-8$y=x2+2x8 positive? We can see from the graph here that unlike our earlier inequality involving lines only, our answer has to be split into two parts,$x<-4$x<4 and$x>2$x>2. No wonder we couldn't solve this algebraically using elementary operations. What if we flipped the inequality sign in the question? So, solve$x^2+2x-8<0$x2+2x8<0. Now we are being asked when the graph$y=x^2+2x-8$y=x2+2x8 is negative. Now our solutions will be$x>-4$x>4 and$x<2$x<2. This can be written as$-44<x$<$<$2$2. ### Solving using Test Values

You may have noticed by now that if a parabola has two distinct roots (when its discriminant is positive), it will always be on one side of the $x$x-axis between the two intercepts, and on the other side of the $x$x-axis outside the two intercepts. Parabolas are either positive between the two intercepts and negative outside of them, or negative between the two intercepts and positive outside of them.