### Video Transcript

In this video, weβll learn how to
solve quadratic equations using an iterative process. Remember, when we solve quadratic
equations, weβre finding the roots of the equation. The solutions to the equation π of
π₯ equals zero correspond to the values where the graph of the quadratic equation π¦
equals π of π₯ intersects the π₯-axis.

But what does it mean to solve an
equation iteratively? Well, iteration is repeatedly
carrying out a process. When solving a quadratic equation
iteratively, we start with an initial value and substitute this into an iterative
formula to obtain a new value. We then repeat this by substituting
this new value back into the formula. The more we do this, the closer we
get to the actual solution to the equation. Essentially, itβs a fancy version
of using trial and improvement. Letβs see what it might look like
using an example.

The equation of the following graph
is π¦ equals negative π₯ squared plus eight π₯ minus 14. Use the iterative formula π₯ sub π
plus one is negative 14 over π₯ sub π minus eight, starting with π₯ naught equals
three, to find the smallest root of the equation negative π₯ squared plus eight π₯
minus 14 equals zero to four decimal places.

Letβs dissect this question. We have a graph of a quadratic
function. Weβre looking to find the smallest
root of the equation negative π₯ squared plus eight π₯ minus 14 equals zero. So, we recall that the root of an
equation π of π₯ equals zero is the value of π₯ where the graph of π¦ equals π of
π₯ intersects the π₯-axis, and itβs simply a solution to our quadratic equation. Now, looking at the graph, we can
estimate the smallest root, the smallest value of π₯, where our graph crosses the
π₯-axis to be around 2.6. But we want a more accurate
result. And so, rather than solving the
equation using something like the quadratic formula or completing the square, we use
iteration.

Weβve been given an iterative
formula in the question, and it does look a little scary with all these
subscripts. Essentially, though, all this means
is that when we substitute a value of π₯ into the formula, the next value of π₯
comes out. So if we substitute π₯ naught into
the equation, we get π₯ one out. Then, substituting π₯ one in gives
us a value of π₯ two and so on. The more we apply this process, the
closer we get to the actual solution. So, letβs see what it looks
like. Our first job is to start with an
initial value of π₯ naught. Sometimes our starting value will
actually be π₯ sub one, but it doesnβt really matter; the process is the same.

According to the question, our
starting value is three. Notice that this is quite close to
the estimate for the root of the equation we initially read from the graph. This will usually be the case. Our next job is to substitute π₯ π
into the formula to get π₯ π plus one out. Well, weβre going to substitute π₯
naught in to begin with, so weβre going to get π₯ one out. π₯ naught is three. So, the first calculation weβre
going to do is negative 14 over three minus eight. That gives us a value of π₯ one
equals 2.8.

Next, we ask ourselves, is π₯ π
equal to π₯ sub π plus one to the required degree of accuracy? Here, thatβs correct to four
decimal places. Well, π₯ naught and π₯ one arenβt
even written to four decimal places, let alone equal to one another. And so we answer no, and we go back
to step two. The moment we can answer yes is
when we know weβre finished. Going back to step two and we see
that substituting π₯ one into our formula will give us our value for π₯ two. We found π₯ one to be 2.8, so we
get π₯ sub two is negative 14 over 2.8 minus eight. Note, though, that if you have a
previous-answer button on your calculator, now is a really good time to use it. Weβre going to be performing this
calculation several times using the previous answer, so itβs a bit of a time
saver.

π₯ sub two is 2.69230 and so
on. Step three, once again, we ask
ourselves, is π₯ one equal to π₯ two? Well, no, theyβre not even correct
to one decimal place. So we go back to step two. We work out the value of π₯ three
by substituting the value of π₯ two into our equation, giving us 2.63768. Once again, comparing these to four
decimal places, we see π₯ two is not equal to π₯ three, and we continue. π₯ four is negative 14 divided by
π₯ three minus eight, giving us 2.61081 and so on. Notice that each time, where
possible, Iβm writing the first five digits after the decimal point. This means I can compare the
answers when theyβre rounded to four decimal places.

Continuing, we get π₯ five is
2.59779, π₯ six is 2.59153, and if we keep going all the way down to π₯ sub 11, we
still answer no to step three. π₯ sub 10 and π₯ sub 11 are indeed
equal to one another to three decimal places but not to four. However, performing the calculation
one more time, that is, negative 14 over the value for π₯ sub 11 minus eight, gives
us 2.58585 and so on. Correct to four decimal places, π₯
sub 11 is 2.5859. And similarly, the five tells us to
round the eight up to a nine, and π₯ sub 12 is also 2.5859 correct to four decimal
places. Weβre now able to answer yes to
step three. And so, we stop here, and we say
that, correct to four decimal places, the smallest root of our equation is π₯ equals
2.5859.

Note, though, we could substitute
this value of π₯ into our original equation. When we do, we get 0.00032119 and
so on. This is very close to the required
value of zero, so we can assume that what weβve done is correct. Now, there is something really
interesting happening here, though. If we were to continue applying the
iterative process, weβd notice that the next few values of π₯ sub π and π plus one
are all equal correct to four decimal places. But theyβre now 2.5858. Essentially, this is just a more
accurate solution to our equation. Of course, following the rules, we
end up with π₯ equals 2.5859, and thatβs absolutely fine.

Letβs have a look at another
example.

The equation of the following graph
is π¦ equals π₯ squared plus three π₯ minus nine. Use the iterative formula π₯ sub π
plus one equals nine over π₯ sub π minus three, starting with π₯ naught equals
negative five, to find the negative root of the equation π₯ squared plus three π₯
minus nine equals zero to four decimal places.

Weβve been given the graph of a
quadratic equation π¦ equals π₯ squared plus three π₯ minus nine. Weβre looking to find the negative
root to the equation π₯ squared plus three π₯ minus nine equals zero. Well, the root is the value of π₯
where the graph intersects the π₯-axis. And that simply gives us a solution
to our quadratic equation. Now, looking at the graph, we can
estimate the negative root to be around negative 4.9. But we want a more accurate
result. And so, rather than, say, applying
the quadratic formula or using completing the square, we use something called
iteration.

Weβve been given an iterative
formula and, while these look a little bit scary, all they mean is that we
substitute a value of π₯ in and the next value of π₯ comes out. So, if we substitute π₯ naught into
our equation, π₯ one comes out. If we substitute π₯ one into the
equation, π₯ two comes out and so on. Now, the more we apply this
process, the more accurate solution we get. Now, letβs recall the steps. The first step is to start with an
initial value of π₯ naught. Now, one way we could find an
initial value of π₯ naught is to find a solution thatβs quite close. So, we said π₯ is approximately
equal to negative 4.9. However, weβre actually told to use
π₯ naught equals negative five, which is of course also close. So weβre going to go with that.

Our next step is to substitute the
value of π₯ π into our iterative formula to find π₯ π plus one. Well, our iterative formula is π₯
π plus one equals nine over π₯ π minus three. So, π₯ one is nine over π₯ naught
minus three. But of course, π₯ naught is
negative five. So we get nine divided by negative
five minus three, which is equal to negative 4.8. Our third step is to ask ourselves,
is the value of π₯ π equal to the value of π₯ π plus one to the given degree of
accuracy? In this case, weβre asking, is π₯
naught equal to π₯ one correct to four decimal places? Well, no. Correct to four decimal places, π₯
naught would be negative 5.0000 and π₯ one would be negative 4.8000. These are clearly not equal.

If we answer no to this question,
we go back to step two. If, however, weβre able to answer
yes, we stop, and we have our solution for π₯. So, letβs go back to step two and
substitute our value for π₯ π into the formula. π₯ two will be nine over the value
for π₯ one minus three. Of course, we saw π₯ one was
negative 4.8. So the calculation weβre going to
do is nine divided by negative 4.8 minus three. At this stage, if you have it, itβs
really useful to use the previous-answer button on your calculator. Weβre going to do nine divided by
the previous answer minus three. And this is because we always
substitute the previous answer into this space in our calculation, so itβll save us
a little bit of time.

π₯ two gives us a value of negative
4.875 or negative 4.8750. π₯ one is not equal to π₯ two, and
so we go back to step two and substitute π₯ two into our formula. If weβre using the previous-answer
button, we can simply press equals. Otherwise, we need to do nine
divided by negative 4.875 minus three. π₯ sub three is negative 4.84615
and so on. Correct to four decimal places,
thatβs negative 4.8462. We see this is not equal to our
previous value for π₯ two, to negative 4.875. And so we continue. Using the previous-answer button,
we should just be able to press equals, and we get negative 4.85714, which, correct
to four decimal places, is negative 4.8571. π₯ three and π₯ four are not equal,
so we continue.

We have to go all the way to π₯ sub
nine and π₯ sub 10 to find two values who round to the same number to the given
degree of accuracy. π₯ sub nine and π₯ sub 10 are both
negative 4.8541 correct to four decimal places. And so, we go back to our flow
chart and we answer yes to step three, and this means we stop. And so, correct to four decimal
places, the negative root to the equation π₯ squared plus three π₯ minus nine equals
zero is π₯ equals negative 4.8541.

Up until this point, weβve been
just given an iterative formula and told to use it. But where does that iterative
formula come from? Well, letβs see.

Show that the equation two π₯
squared minus seven π₯ plus two equals zero can be written in the form π₯ equals
seven over two minus one over π₯. Use the iterative formula π₯ sub π
plus one equals seven over two minus one over π₯ sub π, starting with π₯ naught
equals three, to find a solution to the equation two π₯ squared minus seven π₯ plus
two equals zero to four decimal places.

Letβs begin with the first part of
this question. We want to take a quadratic
equation and write it in the form π₯ equals seven over two minus one over π₯. Now, at first glance, it might look
like weβre making π₯ the subject, but thatβs not really what weβre doing since we
have π₯βs on both sides of our second equation. And so, instead, weβre simply going
to rearrange so we can get it in this form. We take the equation two π₯ squared
minus seven π₯ plus two. At some point, weβre going to need
to divide through by π₯ to achieve this one over π₯. And so, we might infer that at some
point, weβre going to divide this π₯-squared term by π₯. And so that might tell us we need
to leave two π₯ squared on the left-hand side for now.

And so what weβre going to do is
add seven π₯ to both sides of our equation and subtract two. When we do that, weβre left simply
with two π₯ squared on the left-hand side. And on the right, we get seven π₯
minus two. We know we only have one π₯ on the
left-hand side of our equation, so weβre now going to divide through by two. That gives us π₯ squared equals
seven π₯ minus two all divided by two. Now, in fact, what weβre really
going to do next is split this fraction up into seven π₯ over two minus two over
two. And since two divided by two is
equal to one, we get π₯ squared equals seven π₯ over two minus one.

Letβs check how close we are to the
equation required. Well, we have a negative one, and
we have a seven over two, although we have seven π₯ over two. We still have π₯ squared rather
than π₯ on the left-hand side. So now, weβre going to divide
through by π₯. We can do this term by term, and we
get π₯ on left-hand side. And then on the right, we get seven
π₯ over two π₯ minus one over π₯. In this expression seven π₯ over
two π₯, we can divide through by π₯. And so, weβve achieved our
result. We get π₯ equals seven over two
minus one over π₯.

Part two of this question tells us
to use an iterative formula to find a solution to the equation two π₯ squared minus
seven π₯ plus two equals zero. Notice first that our iterative
formula is very similar to the equation we achieved earlier. Instead of π₯βs though, we have π₯
sub π and π₯ sub π plus one. Now, this can seem a little bit
scary, but all that means is that we take a value of π₯, substitute it into our
formula, and we get a second value of π₯ out. So, if we substitute π₯ naught in,
we get π₯ one out. If we substitute π₯ one in, we get
π₯ two out and so on. And the more we do this, the more
accurate our result.

So, letβs begin with π₯ naught
equals three. We said that if we substitute π₯
naught into our formula, weβll get π₯ one out. So π₯ one is seven over two minus
one over π₯ naught or seven over two minus one-third. Thatβs 3.16 recurring. And since weβre looking to give our
answer to four decimal places, we said that π₯ one is 3.1667 correct to four decimal
places. Now, when working with iterative
formulae, we want to find a value of π₯ sub π and π₯ sub π plus one that are equal
to the given number of decimal places, so to four decimal places. We see that π₯ naught and π₯ one
are not equal. So weβre going to substitute our
value for π₯ one into our formula.

π₯ sub two will be seven over two
minus one over our value for π₯ sub one. Thatβs one over 3.16 recurring. And at this stage, itβs much
simpler to use the previous-answer button in our calculator. And this is because weβre going to
perform this calculation a number of times, so it just saves us a little bit of
time. That gives us 3.18421 and so on,
which, correct to four decimal places, is 3.1842. Notice that our previous two values
for π₯, thatβs π₯ sub one and π₯ sub two, are still not equal to four decimal
places. So, weβll do the calculation
again. π₯ sub three is seven over two
minus one over our value for π₯ sub two or seven over two minus one over the
previous answer.

And of course, if weβre using the
previous-answer button, we simply press equals. And we get 3.18595, which is 3.1860
correct to four decimal places. This is not equal to our value for
π₯ sub two, and so we continue. Pressing equals again or
substituting our value for π₯ three back into the formula, we get 3.18612, which,
correct to four decimal places, is 3.1861. This is not equal to our value for
π₯ three correct to four decimal places, so letβs find π₯ five. π₯ five is 3.18613. Correct to four decimal places,
thatβs 3.1861. Notice now that, correct to four
decimal places, π₯ sub four and π₯ sub five are actually equal. And so, we donβt need to continue
with our formula. We stop here. And we say that a solution to the
equation, correct to four decimal places, is π₯ equals 3.1861.

In this video, we learned that
iteration is repeatedly performing a process. When we apply an iterative formula,
we apply the formula over and over again to get closer and closer to the
solution. We saw that the first step is to
start with an initial value of π₯ sub naught. This initial value will usually be
quite close to the actual solution. Note that, very occasionally, step
one will actually be to find an iterative formula. But thatβs very rare.

The next thing we do is we
substitute our value of π₯ π into the formula to get the value of π₯ π plus
one. In other words, when weβre starting
with π₯ naught, we substitute into the formula to get the value of π₯ one. Next, we ask ourselves, is π₯ sub
π equal to π₯ sub π plus one correct to the given degree of accuracy? If not, we go back to step two and
we continue. So, each time we substitute a value
of π₯ in, we get the next value of π₯ out. If, however, the answer is yes, we
stop. And weβve found a solution to our
equation.