### Video Transcript

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.