Physical characteristics of solving an equation

What does it mean to solve an equation, or to find the solution to an equation? We will see that this question can be explored in two important ways:

1. From an algebraic perspective, using the familiar operations to manipulate equations, or
2. From a graphical (or geometric) perspective, using the points of intersection of a function with the coordinate axes and with other functions.

In the 1600s the mathematician and philosopher Rene Descartes made significant contributions to our understanding of the relationship between algebraic and graphical solutions to equations. So let’s walk through this connection, starting with a simple example.

A first walkthrough

Given the equation $x^2=9$x2=9, we might be able to quickly see that it has two solutions: $x=3$x=3 and $x=-3$x=−3. In words, we have asked the question “what number, when squared, gives $9$9?” and we know that the answer is “the numbers $3$3 and $-3$−3”.

In more formal steps we could arrive at the same answer by first taking the square root of both sides of the equation, which would give us $x=\pm\sqrt{9}$x=±√9, and knowing that $3^2=9$32=9 we then have $x=\pm\sqrt{3^2}=\pm3$x=±√32=±3.

That’s the algebraic approach, using only the operations and rules of algebra to solve the equation. Now let’s think like Descartes and view this problem graphically.

The key idea is to relate each side of the equation $x^2=9$x2=9 to a function. So let’s define one function $f(x)=x^2$f(x)=x2 and another function $g(x)=9$g(x)=9. Here are the two functions graphed on the same set of axes.

Notice that the straight line $g(x)=9$g(x)=9 intersects the parabola $f(x)=x^2$f(x)=x2 in two places. The points of intersection have the coordinates $\left(-3,9\right)$(−3,9) and $\left(3,9\right)$(3,9). What is the meaning of these points? It is the $x$x-values of these coordinates that are the solutions to the equation $x^2=9$x2=9.

Looking a bit deeper, we can uncover the algebraic connection. For any input value of $x$x the output of the function $f(x)$f(x) will just be the square of that input value, but the output of $g(x)$g(x) will always be $9$9. In fact, using any input value of $x$x the output of $f(x)$f(x) will be always be different to the output of $g(x)$g(x) except for when $x=\pm3$x=±3. It is only at these two values of $x$x that the output of both functions is the same value. That is, $f(3)=3^2=9$f(3)=32=9 and $g(3)=9$g(3)=9, so $f(3)=g(3)$f(3)=g(3) and similarly $f(-3)=g(-3)$f(−3)=g(−3).

So to wrap it all up, finding the $x$x-coordinates of the points of intersection of the functions $f(x)=x^2$f(x)=x2 and $g(x)=9$g(x)=9 is equivalent to solving the equation $f(x)=g(x)$f(x)=g(x), which is exactly the solution to $x^2=9$x2=9, the algebraic equation that we started with!

Mixing the techniques

There is another way to solve our simple quadratic equation algebraically, and this alternative path will open up a second interesting graphical connection.

Suppose in solving $x^2=9$x2=9 we first chose to subtract $9$9 from both sides of the equation. We get $x^2-9=0$x2−9=0, and we recognise the left hand side as a difference of two squares. This expression can be factorised to give $\left(x+3\right)\left(x-3\right)=0$(x+3)(x−3)=0. By setting each factor equal to $0$0 we have $x+3=0$x+3=0 and $x-3=0$x−3=0, so that the solutions are once again $x=-3$x=−3 and $x=3$x=3.

Now let’s apply Descartes’ big idea to the equation $x^2-9=0$x2−9=0. We relate each side to a function, $f(x)=x^2-9$f(x)=x2−9 and $g(x)=0$g(x)=0, and by finding the points of intersection of the two functions we can solve the equation. Here are the two functions graphed on the same set of axes.

But wait, $g(x)=0$g(x)=0 is just the $x$x-axis! So in this case, finding the solution to $f(x)=g(x)$f(x)=g(x) is the same as finding the $x$x-intercepts of the function $f(x)=x^2-9$f(x)=x2−9.

From simple equations to anywhere

The approach we have explored above can be applied to find the solution to any equation. For example, given the equation $x^2-4x+5=2x+1$x2−4x+5=2x+1, we can either find the intercepts of the two functions $f(x)=x^2-4x+5$f(x)=x2−4x+5 and $g(x)=2x+1$g(x)=2x+1,

or look to solve the equation $x^2-6x+4=0$x2−6x+4=0 by finding the $x$x-intercepts of the function $h(x)=x^2-6x+4$h(x)=x2−6x+4.

 

Examples

QUESTION 1

QUESTION 2

QUESTION 3

Do all equations have solutions?

We have looked at the connection between the solutions to an algebraic equation and the points of intersection of the graphs of functions. If an equation has a solution, then the corresponding graph will have a point of intersection.

But we can think of plenty of pairings of functions that do not have a point of intersection. What does this mean for the solutions to the corresponding algebraic equation? For example, consider the functions $f(x)=x^2$f(x)=x2 and $g(x)=-9$g(x)=−9, graphed below.

Clearly there are no points of intersection between these two functions. And, going backwards, we can see why this is the case. Our corresponding algebraic equation is $x^2=-9$x2=−9. What number, when multiplied by itself, will give a negative value? Remember that the square of any positive real number is positive, and the square of any negative real number is also positive. So there are no real numbers that are the solution to these types of equations.

To summarize this result, an equation that has no real solutions will correspond to a graph where the functions have no points of intersection. And in the same way, if the graphs of two functions do not intersect in the Cartesian plane, then the corresponding algebraic equation has no real solutions.

Although it is beyond the scope of this current topic, it is worth a brief mention that there turns out to be a different type of number - the complex number - that can be used to solve equations like $x^2=-9$x2=−9. And, as you might expect, there is similarly a rich connection between the algebraic solutions to these equations and the geometry of the complex number plane.