Video: Using an Iterative Formula to Find a Root of a Quadratic Equation

The equation of the following graph is 𝑦 = βˆ’π‘₯Β² + 8π‘₯ βˆ’ 14. Use the iterative formula π‘₯_(n + 1) = βˆ’(14/π‘₯_(n) βˆ’ 8), starting with π‘₯_0 = 3, to find the smallest root of the equation βˆ’π‘₯Β² + 8π‘₯ βˆ’ 14 = 0 to 4 decimal places.


Video Transcript

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.

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