Video: Solving Quadratic Equations Iteratively

In this video, we will learn how to solve quadratic equations using an iterative process.

17:53

Video Transcript

In this video, we’ll learn how to solve quadratic equations using an iterative process. Remember, when we solve quadratic equations, we’re finding the roots of the equation. The solutions to the equation 𝑓 of π‘₯ equals zero correspond to the values where the graph of the quadratic equation 𝑦 equals 𝑓 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, it’s a fancy version of using trial and improvement. Let’s see what it might look like using an example.

The equation of the following graph is 𝑦 equals negative π‘₯ squared plus eight π‘₯ minus 14. Use the iterative formula π‘₯ sub 𝑛 plus one is negative 14 over π‘₯ sub 𝑛 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.

Let’s dissect this question. We have a graph of a quadratic function. We’re 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 𝑓 of π‘₯ equals zero is the value of π‘₯ where the graph of 𝑦 equals 𝑓 of π‘₯ intersects the π‘₯-axis, and it’s 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.

We’ve 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, let’s 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 doesn’t 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‘₯ 𝑛 into the formula to get π‘₯ 𝑛 plus one out. Well, we’re going to substitute π‘₯ naught in to begin with, so we’re going to get π‘₯ one out. π‘₯ naught is three. So, the first calculation we’re 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‘₯ 𝑛 equal to π‘₯ sub 𝑛 plus one to the required degree of accuracy? Here, that’s correct to four decimal places. Well, π‘₯ naught and π‘₯ one aren’t 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 we’re 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. We’re going to be performing this calculation several times using the previous answer, so it’s 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, they’re 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, I’m writing the first five digits after the decimal point. This means I can compare the answers when they’re 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. We’re 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 we’ve done is correct. Now, there is something really interesting happening here, though. If we were to continue applying the iterative process, we’d notice that the next few values of π‘₯ sub 𝑛 and 𝑛 plus one are all equal correct to four decimal places. But they’re 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 that’s absolutely fine.

Let’s have a look at another example.

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

We’ve been given the graph of a quadratic equation 𝑦 equals π‘₯ squared plus three π‘₯ minus nine. We’re 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.

We’ve 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, let’s 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 that’s quite close. So, we said π‘₯ is approximately equal to negative 4.9. However, we’re actually told to use π‘₯ naught equals negative five, which is of course also close. So we’re going to go with that.

Our next step is to substitute the value of π‘₯ 𝑛 into our iterative formula to find π‘₯ 𝑛 plus one. Well, our iterative formula is π‘₯ 𝑛 plus one equals nine over π‘₯ 𝑛 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‘₯ 𝑛 equal to the value of π‘₯ 𝑛 plus one to the given degree of accuracy? In this case, we’re 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, we’re able to answer yes, we stop, and we have our solution for π‘₯. So, let’s go back to step two and substitute our value for π‘₯ 𝑛 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 we’re going to do is nine divided by negative 4.8 minus three. At this stage, if you have it, it’s really useful to use the previous-answer button on your calculator. We’re 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 it’ll 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 we’re 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, that’s 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, we’ve been just given an iterative formula and told to use it. But where does that iterative formula come from? Well, let’s 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 𝑛 plus one equals seven over two minus one over π‘₯ sub 𝑛, 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.

Let’s 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 we’re making π‘₯ the subject, but that’s not really what we’re doing since we have π‘₯’s on both sides of our second equation. And so, instead, we’re 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, we’re going to need to divide through by π‘₯ to achieve this one over π‘₯. And so, we might infer that at some point, we’re 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 we’re going to do is add seven π‘₯ to both sides of our equation and subtract two. When we do that, we’re 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 we’re 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 we’re 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.

Let’s 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, we’re 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, we’ve 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‘₯’s though, we have π‘₯ sub 𝑛 and π‘₯ sub 𝑛 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, let’s begin with π‘₯ naught equals three. We said that if we substitute π‘₯ naught into our formula, we’ll get π‘₯ one out. So π‘₯ one is seven over two minus one over π‘₯ naught or seven over two minus one-third. That’s 3.16 recurring. And since we’re 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 𝑛 and π‘₯ sub 𝑛 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 we’re 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. That’s one over 3.16 recurring. And at this stage, it’s much simpler to use the previous-answer button in our calculator. And this is because we’re 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‘₯, that’s π‘₯ sub one and π‘₯ sub two, are still not equal to four decimal places. So, we’ll 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 we’re 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 let’s find π‘₯ five. π‘₯ five is 3.18613. Correct to four decimal places, that’s 3.1861. Notice now that, correct to four decimal places, π‘₯ sub four and π‘₯ sub five are actually equal. And so, we don’t 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 that’s very rare.

The next thing we do is we substitute our value of π‘₯ 𝑛 into the formula to get the value of π‘₯ 𝑛 plus one. In other words, when we’re starting with π‘₯ naught, we substitute into the formula to get the value of π‘₯ one. Next, we ask ourselves, is π‘₯ sub 𝑛 equal to π‘₯ sub 𝑛 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 we’ve found a solution to our equation.

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