### Video Transcript

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

Letβs break this question down. Our graph represents the equation
π¦ equals π₯ squared minus eight π₯ plus eight. This is a quadratic equation. And we know this for two
things. Firstly, the highest power of π₯ is
two; its order is two. So, that tells us we have a
quadratic equation. But also, if we look at the curve,
we have that parabola shape we usually expect.

We want to find the largest root to
the equation π₯ squared minus eight π₯ plus eight equals zero. Notice that all weβve done here is
replace π¦ with zero in our original equation. Now, the roots are the solutions to
these equations when theyβre equal to zero. But what else do they tell us? Well, they show us the value of π₯
where the graph intersects the π₯-axis. And this means the largest root
will be the value of π₯ here.

Now, to me, that looks like itβs
pretty close to π₯ equals 6.9, but we want to be more accurate than this. And so, we use a process called
iteration. This is when we repeatedly apply a
formula to get us closer and closer to our solution. Now, the iterative formula weβve
been given does look a little bit scary. But all it really tells us is we
take a value of π₯, substitute it into our formula, and we get the next value of π₯
back out.

So, if we substitute π₯ naught into
the equation, we get π₯ one out. Substituting π₯ sub one into the
formula gives us π₯ sub two out, and so on. And obviously the more times we do
this, the closer we get to the actual solution. So, letβs clear some space and look
at the steps that we follow.

The first thing we do is we begin
with an initial value of π₯ naught. Very occasionally, we might be
given an initial value of π₯ one, but it really doesnβt matter; the process is the
same. Note also that our starting value
for π₯ β so here, π₯ sub naught β will usually be fairly close to the actual
solution. Remember, we estimated it to be
around 6.9 and weβve been told to use π₯ naught equals seven.

The next thing we do is we take our
value for π₯ naught and we substitute it into the formula to find our value for π₯
one. In general, we take π₯ π,
substitute it into the formula to evaluate π₯ π plus one. In this case then, π₯ one is eight
minus eight over π₯ naught, which is eight minus eight over seven. That gives us a value of 6.85714
and so one.

Then, we ask ourselves, βWell, is
π₯ sub π equal to the value for π₯ sub π plus one to the required degree of
accuracy?β Here, the required degree of
accuracy is four decimal places. And correct to four decimal places,
π₯ one is 6.8571. Itβs quite clear that π₯ naught,
which is seven or 7.0000, is not equal to our value for π₯ one. And so, if we answer no, we go back
to step two. If weβre able to answer yes,
though, we stop and we have our solution.

Of course, we answered no, so weβre
going to go back to step two. That is, weβre going to substitute
π₯ one into our iterative formula, and that will give us a value for π₯ two. And we could do eight minus eight
over 6.85714 and so on. But actually, this is a really good
time to start using the previous-answer button on your calculator. The reason being is that weβre
going to do this process a few times, so it saves us a little bit of time.

Thatβs 6.83333 and so on, which,
correct to four decimal places, is 6.8333. Remember, step three tells us to
compare our value of π₯ π and π₯ π plus one. So now, we compare our value for π₯
sub one and π₯ sub two. We can quite clearly see that π₯
one and π₯ two are not equal correct of four decimal places, so we go back to step
two.

If weβre using the previous-answer
button on our calculator, we simply press equals again. If not, we need to do eight minus
eight over 6.83 recurring. Correct to four decimal places, we
find π₯ three is 6.8293. This is not equal to π₯ two, though
itβs much closer; and so, we continue.

π₯ sub four, by repeating this
process, gives us 6.8286 correct to four decimal places. Thatβs not equal to π₯ three, and
so weβre going to find π₯ five. We have to keep going all the way
to π₯ sub seven. And when we do, we see that π₯ sub
six and π₯ sub seven both round to 6.8284 correct of four decimal places. And so, going back to our flow
chart, weβre able to answer yes to step three. And this means we stop; we have our
solution. Correct to four decimal places, π₯
is equal to 6.8284 is the largest root of the equation π₯ squared minus eight π₯
plus eight equals zero to four decimal places.

Now, of course, we can check our
solution by substituting it back into the equation. When we do, we get negative
0.00015344. Thatβs really close to the answer
of zero. We wouldnβt expect it to be exact
for two reasons. Firstly, weβre using iteration, so
weβre getting closer and closer to the solution. The actual solution will require
almost an infinite number of iterations. Also, we rounded our answer correct
to four decimal places. And so, the fact that itβs very
closest zero, even though itβs not exactly zero, is enough to convince us that we
found the largest root of the equation.