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Quadratic equations - Solving quadratic equations - Edexcel - GCSE Maths Revision - Edexcel - BBC Bitesize

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Learn and revise how to solve quadratic equations by factorising, completing the square and using the quadratic formula with Bitesize GCSE Maths Edexcel.

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DIVIDING IN A GIVEN RATIO

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QUIZ

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A self marking exercise on the application of Pythagoras' Theorem.

Here are some questions which can be answered using Pythagoras' Theorem. You can earn a trophy if you get at least 9 questions correct. Each time you finish a question click the 'Check' button lower down the page to see if you got it right!

Description of Levels

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Level 1 - Finding the hypotenuse

Level 2 - Finding a shorter side

Level 3 - Mixed questions

Level 4 - Pythagoras coordinates

Level 5 - Pythagoras' Theorem exercise

Level 6 - Pythagoras' Theorem harder exercise

Exam Style questions requiring an application of Pythagoras' Theorem and trigonometric ratios to find angles and lengths in right-angled triangles.

Three Dimensions - Three dimensional Pythagoras and trigonometry questions

More on this topic including lesson Starters, visual aids, investigations and self-marking exercises.

Answers to this exercise are available lower down this page when you are logged in to your Transum account. If you don't yet have a Transum subscription one can be very quickly set up if you are a teacher, tutor or parent.

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Curriculum Reference

Pythagoras' Theorem

The area of the square on the hypotenuse of a right angled triangle is equal to the sum of the areas of the squares on the two shorter sides.

You may have learned the theorem using letters to stand for the lengths of the sides. The corners (vertices) of the right-angled triangle is labelled with capital (upper case) letters. The lengths of the sides opposite them are labelled with the corresponding small (lower case) letters.

Alternatively the sides of the right-angled triangle may me named using the capital letters of the two points they span.

As triangle can be labelled in many different ways it is probably best to remember the theorem by momorising the first diagram above.

To find the longest side (hypotenuse) of a right-angled triangle you square the two shorter sides, add together the results and then find the square root of this total.

To find a shorter side of a right-angled triangle you subtract the square of the other shorter side from the square of the hypotenuse and then find the square root of the answer.

Example

AB ² = AC ² - BC ²
AB ² = 4.7 ² - 4.1 ²
AB ² = 22.09 - 16.81
AB ² = 5.28
AB = √5.28
AB = 2.3m (to one decimal place)

The diagrams aren't always the same way round. They could be rotated by any angle.

The right-angled triangles could be long and thin or short and not so thin.

Don't wait until you have finished the exercise before you click on the 'Check' button. Click it often as you work through the questions to see if you are answering them correctly.

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A self marking exercise on the application of Pythagoras' Theorem. Here are some questions which can be answered using Pythagoras' Theorem. You can earn a trophy if you get at least 9 questions correct. Each time you finish a question click the 'Check' button lower down the page to see if you got it right!

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factorising snap

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Box plots can be created from a list of numbers by ordering the numbers and finding the median and lower and upper quartiles.

Draw a box plot for these numbers:

9, 10, 10, 12, 13, 14, 17, 18, 19, 21, 21

The numbers are already in order so the minimum and maximum numbers can be easily read off as 9 and 21, and the middle number can be read as 14.

To find the lower quartile, there are 11 numbers, so \(\frac{11 + 1}{4}\) = 3rd number. The lower quartile will be the 3rd number and the upper quartile is the 9th number.

Drawing these points onto a number line will give the following box plot.

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Box plots - Higher - Representing data - Edexcel - GCSE Maths Revision - Edexcel - BBC Bitesize

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Learn about and revise how to display data on various charts and diagrams with this BBC Bitesize GCSE Maths Edexcel study guide.

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REDUCING BY A PERCENTAGE (basic method)

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OPTIMISATION

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Purplemath

What is linear programming?

Linear programming is the process of taking various linear inequalities (called "constraints") relating to some situation, and finding the best value obtainable under those conditions. A typical example would be taking the limitations of materials and labor, and then determining the optimal production levels for maximal profits under those conditions.

Is linear programming useful in real life?

In real life, linear programming is part of a very important area of mathematics called "optimization techniques".

This field of study (or at least the applied results of it) are used every day in the organization and allocation of resources. These real life systems can have dozens of variables, or hundreds, or more. (For the curious, a history of linear programming: PDF)

In algebra, though, you'll only work with the simple (and graphable) two-variable linear case.

How do you solve a linear programming problem?

The general process for solving linear-programming exercises is as follows:

1. Graph the system of linear inequalities. The area walled off by the various inequalities is called the "feasibility region".

2. Taking the "equals" versions of the inequalities, solve pairs of lines to find the corner points (or "vertices") of the feasibility region.

3. Plug the coordinates of each of the vertices into the model for the situation at hand (that is, into the optimization equation).

4. If you're maximizing, take the corner that gives you the highest value; otherwise, take the one with the lowest value.

5. For word problems, interpret the result within the context given (for example, say "eight meters wide and three meters tall" rather than just "(8, 3)").

Why do you test only the corners?

Somebody figured out that, given a system of linear inequalities and an equation to be optimized (that is, to be maximized or minimized), the system's optimal solution will always be on a corner.

There are complicated explanations for this, but also intuitive ones (for the simple two-variable case that we'll be working with). Either way, the corners are where it's at.

Find the maximal and minimal value of z = 3x + 4y subject to the following constraints:

The three inequalities in the curly braces are the constraints. The area of the plane that they mark off will be the feasibility region. The formula "" is the optimization equation. I need to find the corner points of the feasibility region that return the largest and smallest values of .

My first step is to solve each inequality for the more-easily graphed equivalent forms:

It's easy to graph the system; you take the "equals" version of each inequality (these "equals" versions being the lines forming the edges of the feasibility region), and solve the system of linear equations formed by each pair of lines:

To find the corner points - which aren't always clear from the graph - I'll pair the lines (thus forming a system of linear equations) and solve:

So the corner points - the points I'll be plugging into the optimization equation - are (2, 6), (6, 4), and (−1, −3). To find the solution to this exercise, I only need to plug these three points into "".

As you can see, solving a linear-programming exercise can be a lengthly process, but each individual step is fairly straightforward. Just take your time, label clearly, and work neatly, and you'll get it!

URL: https://www.purplemath.com/modules/linprog.htm

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What is linear programming? What is it used for?

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Given a situation that is modelled by a set of linear inequalities, linear programming is the process of finding the best ('most optimal') solution.

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simplifying QUIZ

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ELIMINATION METHOD

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Equations that have more than one unknown can have an infinite number of solutions. For example, \(2x + y = 10\) could be solved by:

To be able to solve an equation like this, another equation needs to be used alongside it. That way it is possible to find the only pair of values that solve both equations at the same time. These are known as simultaneous equations .

An example of this is:

The unknowns of \(x\) and \(y\) have the same value in both equations. This fact can be used to help solve the two simultaneous equations at the same time and find the values of \(x\) and \(y\) .

Solving simultaneous equations by elimination

The most common method for solving simultaneous equations is the elimination method which means one of the unknowns will be removed from each equation. The remaining unknown can then be calculated. This can be done if the of one of the letters is the same, regardless of sign.

Solve the following simultaneous equations:

First, identify which unknown has the same coefficient. In this example this is the letter \(y\) , which has a coefficient of 1 in each equation.

Either add or subtract the two equations from each other to eliminate the letter \(y\) . In this example the equations will need to be subtracted from each other as \(y - y = 0\) .

If the equations were added together, then \(y + y = 2y\) , and so the letter \(y\) would not be eliminated.

\[\begin{array}{ccccc} 3x & + & y & = & 11 \\ - && - && - \\ 2x & + & y & = & 8 \\ = && = && = \\ x &&& = & 3 \end{array}\]

\[3~\mathbf{\times~3} + y = 11\]

Check the answers by substituting both values into the other original equation. If the equation balances, then the answers are correct:

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Simultaneous equations - Solving simultaneous equations - Edexcel - GCSE Maths Revision - Edexcel - BBC Bitesize

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Equations that have more than one unknown can have an infinite number of solutions. For example, \(2x + y = 10\) could be solved by: To be able to solve an equation like this, another equation needs to be used alongside it.

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Finding the original

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An antique is sold for £550 which is a 10% increase on the price that it was originally bought for. How much was the antique originally bought for?

The original price of the antique is unknown, but no matter what this price was, this is 100% of the value. The item has then been sold for 10% more. This means that it's now 110% and this is worth £550.

Divide both sides by 110 to get 1%:

\[1 \% = 550 \div 110 = 5\]

Multiply both sides by 100 to get 100%:

\[100 \% = 5 \times 100 = \pounds 500 \]

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Reverse percentages - Percentages - Edexcel - GCSE Maths Revision - Edexcel - BBC Bitesize

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Learn about and revise how to calculate percentages to solve real life problems, such as compound interest, with this BBC Bitesize GCSE Edexcel guide.