Previously, we learnt how to solve linear inequalities from a graph.
For instance, if we had the linear inequality $2x-6<0$2x−6<0 we could plot the line $y=2x-6$y=2x−6 like so.
Now solving the inequality $2x-6<0$2x−6<0 just becomes the task of solving $y<0$y<0 for the line $y=2x-6$y=2x−6. In other words, for what values of $x$x is the line below the $x$x-axis?
We can see from the graph that the line is below the $x$x-axis for $x<3$x<3.
So solving linear inequalities just becomes a case of looking where the line is above or below the $x$x-axis. You can use this method for any linear inequality, even ones that don't have zero on the right hand side. This is because you can algebraically rearrange any linear inequality to have zero on the right hand side anyway.
For instance, $-3x-2<-5x+4$−3x−2<−5x+4 can be rearranged like so.
$-3x-2$−3x−2 | $<$< | $-5x+4$−5x+4 | Add $5x$5x to both sides |
$2x-2$2x−2 | $<$< | $4$4 | Subtract $4$4 from both sides |
$2x-6$2x−6 | $<$< | $0$0 |
So actually $-3x-2<-5x+4$−3x−2<−5x+4 is equivalent to the inequality we solved using the graph above, and will have the same solution $x<3$x<3.
You might think that with all the effort it takes to plot a line, we might as well just solve the inequality algebraically by rearranging.
But what if we had a polynomial inequality to solve instead of a linear inequality?
Remember that a polynomial in $x$x is an expression involving terms with coefficients and powers of $x$x.
For example, what if we had to solve $x^3-x^2-2x>0$x3−x2−2x>0?
As you can see, this would be quite tedious and complicated to solve algebraically. Instead, we can graph the polynomial $y=x^3-x^2-2x$y=x3−x2−2x and, as before, try to see where it lies above the $x$x-axis.
To graph the polynomial, we first find it's $y$y-intercept by setting $x=0$x=0.
$y$y | $=$= | $0^3-0^2-2\times0$03−02−2×0 |
$=$= | $0$0 |
So the $y$y-intercept is the origin.
Now we find the $x$x-intercepts by factorising the polynomial and then setting it equal to zero. You can revise how to factorise quadratics here.
$y$y | $=$= | $x^3-x^2-2x$x3−x2−2x | |
$=$= | $x\left(x^2-x-2\right)$x(x2−x−2) | Take out a common factor of $x$x from each term. | |
$=$= | $x\left(x-2\right)\left(x+1\right)$x(x−2)(x+1) | Factorise the quadratic factor into binomials. |
Now we set the polynomial equal to zero to get $x\left(x-2\right)\left(x+1\right)=0$x(x−2)(x+1)=0, which means we have $x$x-intercepts $x=0$x=0,$2$2,$-1$−1.
Also notice that the $x^3$x3 term has a positive coefficient, meaning this graph will be increasing as $x$x goes to infinity.
Using all this information, we can come up with the following graph.
And we can see that the polynomial is above the x-axis between $x=-1$x=−1 and $x=0$x=0, and also when $x>2$x>2.
Remember from our chapter on compound inequalities that we can write "between $x=-1$x=−1 and $x=0$x=0" as $-1
So the solution to the polynomial inequality $-1
Finding this solution algebraically would have been a laborious task. Instead, we used a graph and made the task easier.
As with linear inequalities, we can rearrange any polynomial inequality to get zero on the right hand side, making it a simple task of graphing the polynomial on the left and seeing where it is above or below the $x$x-axis.
Another way we could have solved the above polynomial inequality is by finding the $x$x-intercepts, and using the regions in between them.
$x<-1$x<−1 | $-1 |
$0 |
$x>2$x>2 |
In each region, the polynomial can only ever be always positive or always negative (otherwise it would have to cross the $x$x-axis, which would require another $x$x-intercept).
We choose a test value in each region (usually one that will be easy to do calculations with), and then note down the sign we get, which will tell us the sign of every point in that region.
So for the first region $x<-1$x<−1, we could choose $x=-2$x=−2 and substitute it into the polynomial, giving this.
$y$y | $=$= | $x^3-x^2-2x$x3−x2−2x |
$=$= | $\left(-2\right)^3-\left(-2\right)^2-2\times\left(-2\right)$(−2)3−(−2)2−2×(−2) | |
$=$= | $-8-4-\left(-4\right)$−8−4−(−4) | |
$=$= | $-8-4+4$−8−4+4 | |
$=$= | $-8$−8 |
So the sign for this region must be negative, and we note this in the table.
$x<-1$x<−1 | $-1 |
$0 |
$x>2$x>2 |
$-$− |
We following this same procedure for each region, until we have signs in the table for all of them.
$x<-1$x<−1 | $-1 |
$0 |
$x>2$x>2 |
$-$− | $+$+ | $-$− | $+$+ |
Since the inequality $x^3-x^2-2x>0$x3−x2−2x>0 wants us to find when the polynomial $x^3-x^2-2x$x3−x2−2x is positive, we can just read this off the table.
The polynomial is positive in the region $-1
We want to solve $x^2-x-2>0$x2−x−2>0 by plotting the associated parabola $y=x^2-x-2$y=x2−x−2.
Start by factorising the expression $x^2-x-2$x2−x−2.
Find the $x$x-intercepts of the parabola described by this equation, writing each line of working as an equation. Make sure to write all answers on the same line, separated by commas.
Find the $y$y value of the $y$y-intercept of the curve.
Plot the parabola corresponding to $y=x^2-x-2$y=x2−x−2 on the axes below:
Hence solve the inequality $x^2-x-2>0$x2−x−2>0.
Consider the inequality $\left(4x-3\right)\left(x+6\right)\left(x-7\right)\ge0$(4x−3)(x+6)(x−7)≥0
Solve the corresponding cubic equation: $\left(4x-3\right)\left(x+6\right)\left(x-7\right)=0$(4x−3)(x+6)(x−7)=0
Use the solutions you found in part (a) to solve the inequality. Write your answer using inequalities.
Use your answer from part (b) to express the solution of the inequality in interval notation.
A bus has a velocity that can be positive, negative or zero.
Its velocity is given by $v=5t^2-14t+8$v=5t2−14t+8. For what values of $t$t does the bus have a velocity of $0$0?
Use your answer from part (a) to find the region where the velocity is negative.
On an earlier journey, the velocity of the bus is given by $v=2t^2+2t-40$v=2t2+2t−40. For what values of $t$t does it have a velocity of $0$0? Here, $t$t is allowed to be positive or negative.
Use your answer from part (c) to find the region where the velocity is negative. Here, $t$t is allowed to be positive or negative.