Solve equations numerically: Trial and improvement

﻿Hello, my name is Mr. Clasper and today we are going to be rearranging equations to form iterative formulae. Let's begin this lesson by rearranging some formula to make p the subject. If we take this formula, the first thing we would need to do, would be to add a to both sides, then we would need to divide both sides by seven, giving us p is equal to three k plus a, all over seven. And now p is the subject of our formula. How could we solve this equation? How many different methods are there? Well, we could use the quadratic formula, we could possibly try it to factorise, we could complete the square, we could use trial and improvement or we could attempt to plot it on a graph. Let's rearrange this equation, we could add one to both sides as a first step, then we could subtract five x, we could divide by three, and then we could square root both sides. Taking the same equation, we could rearrange in some different ways. For example, our first step could be to add one to both sides again. However, our next step could be to subtract three x squared this time, and our final step could be to divide through by five or we could as a first step divide through by x, dividing each of the three terms on the left hand side by x, this would give us three x plus five minus 1/x, from here, we could add 1/x to both sides, and my final step could be to subtract five and multiply by 1/3. Remember multiplying by 1/3 is equivalent to dividing by three. Another option could be to add one and subtract five x on both sides, and the final step could be to divide by three x. Notice that in each case, we've left one x on the left hand side of our equation, and also notice that our third example is equivalent to our second example. It's just written in a slightly different way. We can now form iterative formulae from our rearranged equations. In the first example, the subscript of n plus one on the left hand side, refers to our next value for x. So, reading the equation from left to right, this means to find the next value of x, we need to calculate the square root of one minus five lots of the current value of x, all divided by three. In the next example, reading from left to right, this means the next value for x is equal to one minus three lots of the current value of x squared all divided by five, and our last example, reading from left to right means to find the next value for x, we calculate one minus five lots of the current value of x, all divided by three lots of the current value of x. Here are some questions for you to try. Pause the video to complete a task, and click resume once you're finished. And here are your solutions. So for part a, if we subtract two x from both sides, and then square root, we get our final answer for a. And for part b, if we subtract x squared first and then divide this by two, we get our second rearrangement. Here are some questions for you to try. Pause the video to complete your task, and click resume once you're finished. And here are your solutions. So, for the first part, we added five x and subtracted nine from both sides and then we took a square root. And for part b, our first step could have been to divide through by x, this would leave x minus five plus 9/x, and then from here we can rearrange which should give us x is equal to five minus 9/x. Here are some questions for you to try. Pause the video to complete your task, and click resume once you're finished. And here are your solutions. So for part a of your first step was to subtract two x on both sides, and then if we take a cube root, we get the first rearrangement. If we look at part b, your first step was to divide by x, and then from here, you can subtract two and take a square root. And for part c, your first step was to subtract two x and then divide by x squared, leaving x on the left hand side. Here is your last question. Pause the video to complete your task, and click resume once you're finished. And here is your final solution. So, our first step was to divide by x on both sides of our equation, this would leave x squared plus x is equal to 5/x, our next step was to subtract x from both sides, meaning that we have x squared is equal to 5/x minus x, and our last step was to square root, leaving us with a rearrangement from the question. And that brings us to the end of our lesson. So, by now you should be confident rearranging to form iterative formula. Why not give our exit quiz ago just to show off your skills further. I'll hopefully see you soon.

Solve equations numerically: Rearrange to form iterative equations

In this lesson, we will rearrange equations to form iterative equations

Maths

Y11 Maths (Higher)

Key Stage 4

Year 11

Solve equations numerically: Rearrange to form iterative formulae

Solve equations numerically (Iteration)

Algebra (A)

THEME

UNIT

Solve equations numerically: Change of sign

Solve equations numerically: Solving equations using iteration

Lesson overview:Solve equations numerically: Rearrange to form iterative formulae

Core Content

Solve equations numerically (Iteration):