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

The equation of the following graph is π¦ = π₯Β² + 3π₯ β 9. Use the iterative formula π₯_(π + 1) = (9/π₯_(π)) β 3, starting with π₯_0 = β5, to find the negative root of the equation π₯Β² + 3π₯ β 9 = 0 to 4 decimal places.

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Video Transcript

The equation of the following graph is π¦ equals π₯ squared plus three π₯ minus nine. Use the iterative formula π₯ sub π plus one equals nine over π₯ sub π 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.

Weβve been given the graph of a quadratic equation π¦ equals π₯ squared plus three π₯ minus nine. Weβ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.

Weβ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, letβ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 thatβs quite close. So, we said π₯ is approximately equal to negative 4.9. However, weβre actually told to use π₯ naught equals negative five, which is of course also close. So weβre going to go with that.

Our next step is to substitute the value of π₯ π into our iterative formula to find π₯ π plus one. Well, our iterative formula is π₯ π plus one equals nine over π₯ π 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₯ π equal to the value of π₯ π plus one to the given degree of accuracy? In this case, weβ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, weβre able to answer yes, we stop, and we have our solution for π₯. So, letβs go back to step two and substitute our value for π₯ π 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 weβre going to do is nine divided by negative 4.8 minus three. At this stage, if you have it, itβs really useful to use the previous-answer button on your calculator. Weβ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 itβ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 weβ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, thatβ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.