Additional Mathematics for the OCR syllabus - Algebra 7 AS Mathematics Algebra – Solving quadratic equations by completing the square This is the last method of the 4 covered for solving quadratics, & for most pupils the most complicated. It is important that pupils have a good knowledge & understanding, as well as confidence in graphing quadratics, transformations of graphs & the symmetry of quadratics before attempting this lesson. Written by HVaughan (North Chadderton) and LDobson (Blue Coat)

Additional Mathematics for the OCR syllabus - Algebra 7 Objectives Be able to recognise and factorise perfect squares Be able to solve equations by completing the square Again perfect squares have been covered (briefly!) in a previous lesson. If pupils need more confidence with this, you should do some consolidation before proceeding with this lesson! Written by HVaughan (North Chadderton) and LDobson (Blue Coat)

Solving equations by completing the square Additional Mathematics for the OCR syllabus - Algebra 7 Solving equations by completing the square This method for solving quadratic equations involves manipulating the equation so that it includes a perfect square (a + b)2 = a2 + 2ab + b2 Question “ What is a perfect square?” Let’s look at some perfect squares first! Written by HVaughan (North Chadderton) and LDobson (Blue Coat)

Additional Mathematics for the OCR syllabus - Algebra 7 Perfect squares (x + 1)2 = x2 + 2x + 1 (x + 2)2 = x2 + 4x + 4 (x + 3)2 = x2 + 6x + 9 (x + 4)2 = x2 + 8x + 16 (x + 5)2 = x2 + 10x + 25 Halve this no to get … 12, 22, 32, 42 … … Pupils should look for patterns as the next line appears. Questions: “Can you predict the next line?” This should be easy for most! “What’s the connection between the first column and the expanded form?” Written by HVaughan (North Chadderton) and LDobson (Blue Coat)

Additional Mathematics for the OCR syllabus - Algebra 7 Example 1 Concentrate on this bit Solve the equation x2 + 6x = 2 Halve the coefficient of x → (x + 3)2 BUT (x +3)2 = x2 + 6x + 9 Pupils should try to think of a perfect square that starts off the same as the first two terms in the equation. Questions “Why do we concentrate on the first bit, & ignore the rest?” What happens if you concentrate on the ax2 + c instesd?” Subtract 32 i.e. 9 to get x2 + 6x Written by HVaughan (North Chadderton) and LDobson (Blue Coat)

Additional Mathematics for the OCR syllabus - Algebra 7 So [x2 + 6x] = (x +3)2 - 9 Which gives [(x +3)2 – 9] = 2 (x +3)2 = 2 + 9 → Square root, remember 2 solutions √ (x +3)2 = 11 x +3 = √11 The example is revealed line by line. Students should try to predict the next line each time. Discuss different ways of recording the solution. How can you check the answer? How many roots should there be? x = -3 √11 x = -6.32 or x = 0.32 (2 d.p.) Written by HVaughan (North Chadderton) and LDobson (Blue Coat)

Additional Mathematics for the OCR syllabus - Algebra 7 Example 2 Solve the equation x2 - 8x - 6 = 0 x2 - 8x = 6 Halve the coefficient of x → (x - 4)2 Subtract 42 (x - 4)2 – 16 = 6 (x - 4)2 = 6 + 16 (x - 4)2 = 22 The example is revealed line by line. Students should try to predict the next line each time. Discuss different ways of recording the solution. How can you check the answer? How many roots should there be? x - 4 = √22 x = 4 √22 x = -0.69 or x = 8.69 (2 d.p.) Written by HVaughan (North Chadderton) and LDobson (Blue Coat)

Additional Mathematics for the OCR syllabus - Algebra 7 How can we find the equation of the line of symmetry of an equation using the “completed square” form? Look at y = [x2 + 6x] -2 (see example 1) y = [(x + 3)2 - 9] - 2 y = (x + 3)2 - 11 You might want to miss this last bit out, depending on the ability of your class. They will need graphical calculators or a graphing package. Pupils should look back at example 1 from this lesson to work through this bit. You need a graphical calculator or a graphing package!! Written by HVaughan (North Chadderton) and LDobson (Blue Coat)

Additional Mathematics for the OCR syllabus - Algebra 7 Start by drawing y = x2 y = x2 Ask the question “How can you get closer to the equation of the curve we want starting form this basic curve?” Pupils may need to be reminded about transformations of graphs. Root x = 0 y intercept, y = 0 Line of symmetry at x = 0 Written by HVaughan (North Chadderton) and LDobson (Blue Coat)

Additional Mathematics for the OCR syllabus - Algebra 7 Translate to get 3 places to the left to get y = (x + 3)2 Additional Mathematics for the OCR syllabus - Algebra 7 y = (x + 3)2 y = x2 y intercept, y = 9 This should have been covered previously in mathematics lessons. Remind students that the graph is translated BACK 3 spaces to find f(x+3) – often they expect it to move the other way. Root x = -3 Line of symmetry at x = -3 Written by HVaughan (North Chadderton) and LDobson (Blue Coat)

Additional Mathematics for the OCR syllabus - Algebra 7 Translate 11 places down to get y = (x – 3)2 - 11 y = (x + 3)2 Line of symmetry at x = -3 y = x2 y = (x + 3)2 -11 Translate the new curve 11 spaces down to get to our finishing point. Alternatively, draw y = 11 on he graph and read off the intersections with f(x + 3) this will give an equivalent answer. (Remind students of the work in lesson A4) on graphical solutions of quadratic equations. Root x = -6.32 Root x = 0.32 y intercept, y = -2 Minimum point at (-3, -11) Written by HVaughan (North Chadderton) and LDobson (Blue Coat)

Additional Mathematics for the OCR syllabus - Algebra 7 Repeat this method to find the line of symmetry and ii) the minimum point of y = x2 – 8x – 6 Extra practice on this topic would be beneficial. Use this question &/or see the Additional mathematics for OCR textbooks, or any A’ level textbook for questions. For extension try some curves that start with say 4x2 + ….. or 2x2 + ….. Or x2 + 5x + … Etc. Written by HVaughan (North Chadderton) and LDobson (Blue Coat)