Question Video: Use an Iterative Formula to Find a Root of a Quadratic Equation Mathematics

The equation of the graph is 𝑦 = π‘₯Β² βˆ’ 8π‘₯ + 8. Use the iterative formula π‘₯_(𝑛 + 1) = 8 βˆ’ (8/π‘₯_(𝑛)), starting with π‘₯β‚€ = 7, to find the largest root of the equation π‘₯Β² βˆ’ 8π‘₯ + 8 = 0 to 4 decimal places.

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Video Transcript

The equation of the following graph is 𝑦 equals π‘₯ squared minus eight π‘₯ plus eight. Use the iterative formula π‘₯ sub 𝑛 plus one equals eight minus eight over π‘₯ sub 𝑛 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.

Let’s break this question down. Our graph represents the equation 𝑦 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 we’ve done here is replace 𝑦 with zero in our original equation. Now, the roots are the solutions to these equations when they’re 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 it’s 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 we’ve 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, let’s 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 doesn’t 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 we’ve 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‘₯ 𝑛, substitute it into the formula to evaluate π‘₯ 𝑛 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, β€œWell, is π‘₯ sub 𝑛 equal to the value for π‘₯ sub 𝑛 plus one to the required degree of accuracy?” Here, the required degree of accuracy is four decimal places. And correct to four decimal places, π‘₯ one is 6.8571. It’s 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 we’re able to answer yes, though, we stop and we have our solution.

Of course, we answered no, so we’re going to go back to step two. That is, we’re 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 we’re going to do this process a few times, so it saves us a little bit of time.

That’s 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‘₯ 𝑛 and π‘₯ 𝑛 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 we’re 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 it’s much closer; and so, we continue.

π‘₯ sub four, by repeating this process, gives us 6.8286 correct to four decimal places. That’s not equal to π‘₯ three, and so we’re 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, we’re 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. That’s really close to the answer of zero. We wouldn’t expect it to be exact for two reasons. Firstly, we’re using iteration, so we’re 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 it’s very closest zero, even though it’s not exactly zero, is enough to convince us that we found the largest root of the equation.

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