In this video we’re gonna talk about finding the roots of a quadratic
equation. Specifically we’re gonna look at how to find them from a graph. First though we need to remember that a quadratic equation is an equation in
the form 𝑦 equals 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐. We sometimes call it a polynomial of
order two. So it has an 𝑥 squared term — something times 𝑥
squared, so just a number times 𝑥 squared — it has an 𝑥 term,
so some number times 𝑥, and it has a constant term which is just a number and that
could be positive or negative.
Now remember the multiplier of 𝑥, so 𝑏 in our
equation, or the constant term on its own, 𝑐 in our equation or even both could
actually turn out to be zero. So for example, we could have something like five 𝑥 squared plus three 𝑥 plus
two. And in that case the 𝑎-value would be positive five, the 𝑏-value
will be positive three, and the 𝑐-value will be positive two.
Or we could have an equation like 𝑦 equals two 𝑥 squared minus nine, in
which case the 𝑎-value will be two, the 𝑏-value will be zero — so
there’s no 𝑥 term — and 𝑐 in this case is negative nine.
Another example 𝑦 equals three 𝑥 squared plus seven 𝑥. And in that case the
𝑎-value is three,
𝑏 would be positive seven, and 𝑐 would be zero. So there’s no constant
term on the end.
And another example is just 𝑦 equals 𝑥 squared, in which case the
𝑎-value will be one, the 𝑏-value will be zero, and the 𝑐-value is
also zero. So these are all examples of quadratic equations. And it’s okay for
𝑏 to be zero; it’s okay for 𝑐 to be zero. But notice that 𝑎 is
never zero in a quadratic equation because then it wouldn’t be quadratic; we wouldn’t have an
𝑥 squared term.
So what is a root? And why would we want to find it? Well remember that there
are lots of real-life situations that can be modelled using quadratic equations. And in lots of
cases it would be really useful to know what the 𝑥-coordinate is when the
𝑦-coordinate is zero. So for example, what time will the asteroid hit the ground?
When will the furnace run out of fuel? When will the chemical reaction run out of catalyst? Or
when will the car stop skidding? So finding the roots of quadratics can be a really useful
thing to do. And so all the root is is what is the 𝑥-coordinate when a
𝑦-coordinate is zero. In other words at what point does that curve cut through
Now there are lots of different ways to work out what the roots are. So there
are some algebraic ways, which are probably the best ways. But in this video we’re just gonna
concentrate on reading those values off of a graph that we’ve been given. So let’s have a look
at some examples.
Right number one, so we’re given the graph of 𝑦 equals 𝑥 squared minus
three 𝑥 minus four. And the question asks using the graph shown, find the roots of the quadra-
quadratic 𝑦 equals 𝑥 squared minus three 𝑥 minus four. Well that’s great because the
quadratic we’re trying to find the roots of is the one that we’ve been given the graph for. So
all we’ve gotta do is see where does it cut the 𝑥-axis and then we just have to
read off those 𝑥-coordinates. So this graph cuts the 𝑥-axis in two places: one when 𝑥 is
equal to negative one and one when 𝑥 is equal to four. So that’s our answer.
There are two answers: 𝑥 equals negative one and 𝑥 equals four.
So number two, we’re given a graph and the question says using the graph given
find the roots of the quadratic equation 𝑦 equals negative 𝑥 squared plus 𝑥 plus
six. So the first thing you’ll notice is the equation it shows on the graph looks
slightly different at first glance than the equation that we’re given in the question. But remember
we need to look at the coefficient of the 𝑥 squared term, the coefficient of
the 𝑥 term, and then the constant term. So if we do that for the question, we see
that the 𝑎-value is negative one, the 𝑏-value is one, and 𝑐 is
six. And although they’re in a different order on the graph, we’ve still got an
𝑎-value of negative one, a 𝑏-value of one, and a 𝑐-value of
six. So we’ve got the same equation. So the graph matches the equation in the question. So to find the roots, we
just need to work out what’s the 𝑥-coordinate, where the graph cuts the
𝑥-axis. So we see the curve cuts the 𝑥-axis in two places: one is where
𝑥 is equal to negative two and one is where 𝑥 is equal to positive three.
So that’s our solution.
Right so for number three, we’re given a graph and then the question says using
the graph shown, find the roots of the equation 𝑦 equals 𝑥 squared minus six 𝑥 plus
nine. So just checking the equation we’re given in the question matches the
equation we’re given in the graph. So it’s just a matter of reading off where about that cuts
the 𝑥-axis. Now this question is slightly different to the others because in fact it
doesn’t cut the 𝑥-axis as such; it just touches it in this one place at
𝑥 equals three.
Now we’re gonna touch on algebraic solutions to these questions in another
video. But let’s just rearrange 𝑦 equals 𝑥 squared minus six 𝑥 plus nine. And you can see
there that if I factor it I’ve got 𝑦 equals 𝑥 minus three times 𝑥 minus three. And the definition of a root is the 𝑥-value that generates a
𝑦-coordinate of zero. So if I put my 𝑥 minus three times 𝑥 minus three and
then put 𝑦 equals zero, we can see that something times something or something
times itself is equal to zero. Now in order for that to be true at least one of those
things has to be zero either 𝑥 minus three is zero or 𝑥 minus three is zero. So the
same thing is zero and each one of these generates a root.
Then in the first case obviously that’s 𝑥 equals three, we’ll make
that bracket equal to zero. And the same goes for the other — the contents of the other bracket. So we just happen to have a situation here where we’ve got two answers, but they
happen to be the same. And we call that “repeated roots.” So in situations like this where the graph just touches the 𝑥-axis,
the answer would be that we have repeated roots at 𝑥 equals three.
Moving on to number four then, so we’re given a graph and then it says using
the graph find the roots of 𝑦 equals negative 𝑥 squared plus four 𝑥 take away two. And
that equation is the same as the equation of the graph that we’ve been given. So it’s just a
matter of reading off the 𝑥-coordinates of the points where it cuts the
𝑥-axis. But now we’re running into a little bit of difficulty. In all the previous
questions, it’s done that exact coordinate point — so whole number solutions, integer solutions
for 𝑥. But this now we’ve got a solution at just above nought point five and a
solution just below three point five. So I would say it looks like it’s about nought point six and three point
So this highlights the great limitation of the graphing method because we can
only really get approximate solutions. We can’t get really really super accurate solutions to
these because we’re reading values from a graph. And we’ve only got the scale of the graph to
go from; so we can’t get really really super accurate answers. If I did this algebraically, the actual
answers are 𝑥 equals two minus root two and 𝑥 equals two plus root
two. So with exact answers I can pick a number of decimal places. And I can give
the answer as accurately as I like. Reading from the graph, we’d probably start to one or maybe
two decimal places depending on the scale of the graph.
Now with number five, we’re given a graph and we’re told to use that graph to
find the roots of the equation 𝑦 equals 𝑥 squared minus eight 𝑥 plus eighteen. So we’ve got the
same equation in each case. So we should just be able to see where that graph cuts the
𝑥-axis and read off the 𝑥-coordinates. Now remember quadratic curves are parabolae. So this is gonna go on for
infinity in that direction and on for infinity in that direction. The curve turns around at this
point here, but it never actually quite touches or reaches the 𝑥-axis. So there
aren’t any roots for this particular graph or for this particular equation.
So what this means is that for this particular equation, it doesn’t matter
what value you replace 𝑥 with. You’re never gonna come up with a number which is
gonna generate a 𝑦-coordinate of zero. So the answer might be puzzling a little bit. We call it “no real roots”; you
could just say there are no roots. But in fact if we invent an entirely new number system —
complex numbers or imaginary numbers — it is possible to use some of these complex or imaginary
values for 𝑥, which would generate a 𝑦-coordinate of zero. But that’s
a story for a completely different day for now. All you need to think about it is with these
situations if the curve doesn’t cut the 𝑥-axis. We just say there are no real
So the last question we’re gonna look at is this one. Now this one gets a
little bit tricky towards the end, but we’ll worry about that when we get there. So we’re given
the graph of 𝑦 equals 𝑥 squared minus two 𝑥. And we need to, in part a,
use that graph to find the roots of 𝑦 equals 𝑥 squared minus two 𝑥.
So the equations are the same. So it’s just a matter of looking at the graph and finding out
where it cuts the 𝑥-axis and looking at those 𝑥-coordinates. So the answers there are 𝑥 equals zero and 𝑥 equals two.
Now part b, use the graph to solve 𝑥 squared minus two 𝑥 equals one. Now this isn’t
finding the roots as such, but we’re trying to find which 𝑥-coordinates are
going to generate a 𝑦-coordinate of one. So when we were
looking for the roots, we were trying to find out which 𝑥-coordinates generate a
𝑦-coordinate of zero. But in this case we’re looking for a 𝑦-coordinate
of one. So if I draw the line 𝑦 equals one on my graph, wherever that
graph cuts through 𝑦 equals one that’s gonna generate my 𝑥-coordinates
for me. So there we have it. I’ve plotted 𝑦 equals one and our original
graph cuts through 𝑦 equals one in these two points. And if I read off the
𝑥-coordinates in the first case on the left here, I’ve got minus nought point four. And for this right-hand case here, that looks like that’s 𝑥 equals two
point four. So again we’ve discovered a bit of a limitation of using graphs because the
completely hundred percent accurate answers would’ve been 𝑥 is one minus root two and
𝑥 is one plus root two. But using the graphs, what we’ve got was as close as we can get.
Now for part c, it’s all got a little bit tricky because the problem is that the
equation we were given in the graph 𝑦 equals 𝑥 squared minus two 𝑥 isn’t the same as
the equation that we’re trying to solve here. We haven’t got 𝑦 equals 𝑥 squared minus
two 𝑥. We’ve got on the left-hand side- we’ve got 𝑦 equals 𝑥 squared minus three 𝑥 plus
one. And we’re putting the 𝑦-coordinate equal to zero. So what we ideally
have is the quadratic of 𝑦 equals 𝑥 squared minus three 𝑥 plus one. And we’ll just be
looking for where that cuts the 𝑥-axis and has a 𝑦-coordinate of
So what we need to do is to say well what steps do we need to go through in
order to get this-this expression to look like the expression the we’ve-that we’ve got for the
graph. Well we could do that in a few different stages. So to get from negative three 𝑥 up to negative two 𝑥. I need
to add 𝑥. So if I add 𝑥 to both sides of that equation, I’ve got this. And now if I subtract one from each side of the equation, I’ll
end up with just 𝑥 squared minus two 𝑥 on the left-hand side. Now just like in part b, we had to plot 𝑦 equals 𝑥 squared minus
two 𝑥 and 𝑦 equals one and see where they cross over. Now I’m gonna plot 𝑦 equals 𝑥 squared minus two 𝑥 and 𝑦 equals 𝑥 minus one
and see where they cross over — in other words where their 𝑦-coordinates would be
equal. So 𝑦 equals 𝑥 minus one. I could write that as 𝑦 equals one 𝑥
And if you remember from equations of straight lines, the one in front of the
𝑥 — the multiplier of the 𝑥 — tells us the slope of that straight line.
And the negative one tells us where it cuts the 𝑦-axis — the 𝑦
intercept. So I’ve got negative one is where it cuts the 𝑦-axis and the slope is
one. So every time I increase my 𝑥-coordinate by one, my 𝑦-coordinate
is also going up by one. So this is gonna be on the point on the line as well. Increase my
𝑥-coordinate by one, the 𝑦-coordinate is gonna go up by one. So I
can draw my straight line 𝑦 equals 𝑥 minus one through those points. And then I need to work out where does that quadratic graph cut this straight
line. So that’s here and here; that’s when though the line 𝑦 equals 𝑥 one has the
same 𝑦-coordinate as the quadratic graph of 𝑦 equals 𝑥 squared minus
So now let’s read off the 𝑥-coordinates where they cross over.
So the left-hand point here has got an 𝑥-coordinate of nought point four and the right-hand point has got an 𝑥-coordinate of- looks like
two point six. And if I did that algebraically, the actual answers will be three minus root
five over two or three plus root five over two. And they’re pretty close to nought point four and two
point six. But again it just shows the limitations of doing your working out using the graph.
Reading values from a graph rather than algebraically, you can’t get quite such accurate
So let’s just do a summary then. The roots of a quadratic equation are the
𝑥-coordinates of the points on the graph that have 𝑦-coordinates
of zero, so the 𝑥-values in the equation that generate a 𝑦-value of zero — in
other words the points where it cuts the 𝑥-axis. So on a graph you can just read the 𝑥-coordinates from the graph
where the curve cuts the 𝑥-axis. And these might not be whole numbers; so you may only be able to give
approximate solutions. And there could be two or one or even no real solutions. So good luck with finding roots of quadratic equations from graphs.