### Video Transcript

Prove that the equation four plus six π₯ squared minus two π₯ cubed is equal to zero
can be rearranged into the form π₯ is equal to two over π₯ squared plus three.

To answer this question, letβs start by looking at the rearranged form. Notice how the numbers in the equation have changed. We no longer see a four or a six at all. We can see though that each of these numbers shares a common factor of two. And we can, therefore, divide the entire equation by two. That gives us two plus three π₯ squared minus π₯ cubed is equal to zero.

Already, itβs starting to look a little bit more like what weβre trying to get
to. We now have a two and a three in each equation. We can see that the two in our rearranged form is being divided by π₯ squared. Letβs divide the whole equation by π₯ squared and see what happens.

Two divided by π₯ squared is simply two over π₯ squared. Three π₯ squared divided by π₯ squared is three. And π₯ cubed divided by π₯ squared is just π₯. We know that because of this rule: when we have two numbers that have indices and
weβre choosing to divide them, as long as the base is the same, in that case, the
base is the letter π₯, we can subtract the powers. That means then when we divide π₯ to the power of three by π₯ squared, three minus
two is one. That gives us π₯ to the power of one or simply π₯.

Now, we can see that weβre getting even closer to our rearranged form. Three is now our constant and we have a single π₯, which we also have in the
rearranged form. π₯ is the subject in our rearranged equation. So weβre going to add π₯ to both sides. That gives us two over π₯ squared plus three is equal to π₯. Since the equal sign means the same, we can swap whatβs on both sides of the equation
to give us π₯ is equal to two over π₯ squared plus three.

Starting at π₯ nought equals three and using π₯ π plus one is equal to two over π₯
π squared plus three, find the values of π₯ one, π₯ two, and π₯ three.

This is an example of an iterative formula. It looks much scarier than it is. All itβs saying is that to find the next value of π₯, you take the current value and
substitute it into the formula. The starting value π₯ nought is three. In this case then, to find π₯ one, weβll substitute our value of π₯ nought into the
formula. That gives us two divided by three squared plus three. Typing this into our calculator gives us 3.2 recurring.

Then to find the value of π₯ two, we substitute π₯ one into our formula. Itβs helpful to use the previous answer button on your calculator to save time in the
next few steps and to prevent any errors from rounding too early. Two divided by 3.2 recurring squared plus three is equal to 3.1926.

Finally, to get π₯ three, we substitute the value of π₯ two into our formula or if we
use the previous answer button to get π₯ two, we can simply press equals again to
get the value of π₯ three. Thatβs 3.1962.

At this point, we need to choose a suitable level of accuracy for our answer as this
has not been specified in the question. A minimum of three decimal points will do it. 3.2 recurring is already in exact form. Rounding 3.1926 to three decimal places gives us 3.193. Similarly, rounding the value of π₯ three to three decimal places gives us 3.196.

If you continue this process, the values of π₯ nine, π₯ 10, and π₯ 11 are all equal
to 3.19582 when rounded to five decimal places. Explain what this number represents.

We can see that our answers for π₯ one, π₯ two, and π₯ three are all tending to this
number of 3.19582. Thatβs to say theyβre getting closer and closer to it each time. In fact, an iterative formula is used to approximate a solution to an equation. In this case, the value 3.19582 is an approximation to the solution of the original
equation we rearranged: four plus six π₯ squared minus two π₯ cubed equals zero. If we were to substitute this number back into that equation, we get an answer
extremely close to zero.