Sandringham School Sixth Form. AS Maths. Bridging the gap

Transcription

1 Sandringham School Sixth Form AS Maths Bridging the gap

2 Section 1 - Factorising be able to factorise simple expressions be able to factorise quadratics The expression 4x + 8 can be written in factor form, the numbers 4 and x + 2 are the factors. The process of writing 4x + 8 in the form 4(x + 2) is called factorising. Example 1A Factorise (a) 6x + 10, (b) 2y + 12 (c) 5t 15 (d) 6a + 18 (e) ab + a 2 b (a) 6x (3x + 5) The number 2 divides into both terms of 6x + 10 (b) 2y (y 6) Notice the sign on both sides. (c) 5t 15 5(t + 3) Take care with the signs. You could also write 5( - t 3) (d) 6a (3a + 9) 6(a + 3) 2(3a + 9) is not completely factorised because (3a + 9) can be factorised as 3(a +3). (e) ab + a 2 b ab(1 a) a 2 b ab ab(a 1) Rearrange left hand side to move negative. ab divides both terms. Now have a go at section 1A questions 1,3,6,7,9 and 10. If the signs in a quadratic, for example x 2 + 7x + 10, are both positive, then the factors are of the form (x + )(x + ). If the sign of the constant term is positive, but the term involving x is negative, for example x 2 7x + 10,then the factors are of the form (x )(x ). If the sign of the constant term is negative, for example x 2 7x 10, or x 2 + 7x 10, then the factors are of the form (x )(x + ).

3 Example 1B Factorise (a) x 2 8x + 12, (b) x 2 + 3x 4, (c) x 2 7x 18 (x (a) )(x x 2 ). 8x + 12, x 2 8x + 12 (x 2)(x 6) The factors have the form (x -...) (x -...). 12 can be factorised as 1 x 12, 2 x 6, and 3 x 4. The possibilities are (x - 1) (x -12), (x - 2) (x -16) and (x - 3) (x - 4) Check which is correct. (b) x 2 + 3x 4, x 2 + 3x 4 (x 1)(x + 4) The factors have the form (x -...) (x +...). 4 can be factorised as 1 x 4, and 2 x 2. The possibilities are (x - 1) (x + 4), (x + 1) (x - 4), and (x - 2) (x + 2) Check which is correct. (c) x 2 7x 18, x 2 7x 18 (x 9)(x + 2) The factors have the form (x -...) (x +...). 18 can be factorised as 1 x 18, 2 x 9 and 3 x 6. The possibilities are (x - 1) (x + 18), (x - 18) (x + 1), (x - 2) (x + 9), (x - 9) (x + 2), (x - 3) (x + 6), (x - 6) (x + 3), Check which is correct. In practice, you would probably do much of the work in these examples in your head or on rough paper, and all you might write down is the very last line; the examples show the process in some detail. Now have a go at section 1B, questions, 11, 12, 14, 16, 17, 22, 23, 24 and 25.

4 Section 2 - The difference of two squares be able to apply the difference of two squares rule. Quadratics like x 2 16 and 100 4x 2 have no term involving x, and both the other terms are squares separated by a minus sign. They are referred to as the difference of two squares. Difference of two squares rule: x 2 k 2 (x + k)(x k) Example 2 Factorise (a) x 2 9, (b) 4x 2 81, (c) 25x 2 y 2 (d) (3y + 1) 2 (y + 2) 2 (a) x 2 9 (x + 3)(x 3) (b) 4x 2 81 (2x + 9)(2x 9) (c) 25x 2 y 2 (5x + y)(5x y) (d) (3y + 1) 2 (y + 2) 2 {(3y + 1) + (y + 2)}{(3y + 1) (y + 2)} {3y y + 2}{3y + 1 y 2} (4y + 3)(2y + 1) Use x 2 k 2 (x + k) (x k) with k 2 4x 2 81 is (2 x) Apply the difference of two squares rule. 25x 2 y 2 is (5 x) 2 y 2. Apply the difference of two squares rule. Use x 2 k 2 (x + k) (x k) with x 3y+ 1 and k y + 2. Then expand the inside brackets. Finally simplify the brackets. Now have a go at section 2,questions 1, 3, 4, 5 and 9.

5 Section 3 Changing the subject. be able to make x the subject of a formula A formula is a mathematical equation containing two or more letters, for example A(px + q) B The process for changing the subject of a formula is very similar to that for solving an equation. If there are brackets in it, then, just as with equations you should expand them first. Examples 1 In each case, make x the subject of the formula. (a) ax + a b + c (b) x+p p p q + q p (c) x2 a 2 y2 b 2 1 (a) ax + a b + c a + ax a b + c a ax b + c a x b+c a a Get the terms with x on one side, and everything else on the other. Do this by subtracting a from both sides. Divide through by a (b) x+p p p q + q p q(x + p) p 2 + q 2 qx + pq p 2 + q 2 qx p 2 + q 2 pq x p2 +q 2 pq q Recall that the fraction bar acts as a bracket, so put brackets in. Then multiply both sides by pq and cancel. Expand the bracket. Subtract pq from both sides. Divide both sides by q.

6 (c) x 2 a 2 y2 b 2 1 b 2 x 2 a 2 y 2 a 2 b 2 b 2 x 2 a 2 b 2 + a 2 y 2 x 2 a2 b 2 + a 2 y 2 b 2 x ± a2 b 2 + a 2 y 2 b 2 Clear the fractions by multiplying both sides by a 2 b 2. Then solve the equation for x 2. Finally, take the square root: the sign + means plus or minus, showing that there are two possible values of x. Now have a go at section 3, questions 1, 2, 3, 6, 7, 9, 10, 12, 13 and 15. Section 4 Simplifying Algebraic Fractions. be able to simplify algebraic fractions by cancelling. Start by factorising the numerator and denominator if necessary. Then you can divide the numerator by any common factor to get the simplified fraction. Example Simplify 3x 2 8x + 4 6x 2 7x + 2 3x 2 8x + 4 6x 2 7x + 2 (x 2)(3x 2) (2x 1)(3x 2) x 2 2x 1 Factorise numerator and denominator Cancel common factors. Now have a go at section 4, questions 1, 3, 5, 7 and 9.

7 Section 5 Combining algebraic fractions be able to combine numbers and algebraic fractions be able to handle fractions that include brackets. Important rules to remember The multiplication rule for fractions is The division rule for fractions is a b c a c d b d ac bd a b c d a b d c ad bc The addition/ subtraction rule for fractions is a b ± c a d ± c b ad ± cb d bd bd Examples In each case express as a single fraction (a) 5 + x+2 x 1 (b) 2 + x+1 x 2 x 3 (c) x x 1 (x+3) 2 (x+3) (d) x2 6x 7 x 2 4 x2 x 2 x 2 7x (e) x 2 4 x 2 +3x+2 (x 2)2 x 2 +2x+1 (a) 5 + x+2 5(x 1)+(x 2) x 1 (x 1) 5x 5 + x 2 (x 1) 6x 7 x 1 The number 5 is the fraction 5 1. Multiply out brackets. Collect like terms.

8 (b) 2 + x+1 2(x 3)+(x+1)(x 2) x 2 x 3 (x 2)(x 3) 2x 6 + x2 x 2 (x 2)(x 3) x2 + x 8 (x 2)(x 3) The denominator is (x 2)(x 3) Multiply out brackets. Collect like terms. (c) x x 1 x(x+3) (x 1) (x+3) 2 (x+3) (x+3) 2 x2 + 3x x + 1 (x + 3) 2 x2 + 2x + 1 (x + 3) 2 (x + 1)2 (x + 3) 2 The denominator is (x + 3) 2 Multiply out brackets. Collect like terms. Factorise the top. (d) x2 6x 7 x 2 4 x2 x 2 x 2 7x (x+1)(x 7) (x+1)(x 2) (x 2)(x+2) (x)(x 7) (x + 1)2 x(x + 2) Factorise numerator and denominator where appropriate. Cancel common factors and apply rules for multiplying fractions. (e) x 2 4 (x 2)2 (x+2)(x 2) (x 2)2 x 2 +3x+2 x 2 +2x+1 (x+2)(x+1) (x+1) 2 (x + 2)(x 2) (x + 2)(x + 1) (x + 1) (x + 2) (x + 1)2 (x 2) 2 Factorise numerator and denominator where appropriate and apply rules for dividing fractions. Cancel common factors. Now have a go at section 5, questions 1, 3, 5, 7, 9, 11, 12, 18 and 19.

9 Section 6 Simultaneous equations Be able to solve simultaneous equations by both elimination and substitution. Be able to solve simultaneous equations where both equations are linear. Be able to solve simultaneous equations where one equation is linear and the other quadratic. Elimination In each case you need to reduce the equations to one involving x or y alone. This is called eliminating x or y. If you can t add or subtract to eliminate one of the letters straight away, chose the letter you want to eliminate and multiply one or both of the equations first to make the coefficients equal. Then you can eliminate that unknown. You may have to rearrange one of the equations so they are of a similar form. Example Solve the simultaneous equations 2x + 3y 1 7x y 2x + 3y 1 (1) 7x 5y 12 (2) 14x + 21y 7 (3) 14x 10y 24 (4) 31y 31 y 1 2x + 3 ( 1) 1 2x 3 1 2x 2 x 1. Rearrange the equations into a similar form. Multiply equation (1) by 7 and (2) by 2 to make the coefficients of x the same. Subtract (4) from (3) to eliminate x and solve to find y. Substitute y into a simpler equation and solve to find x. The solution is x 1, y 1.

10 Substitution This method is useful when one of the equations is a quadratic, but it can be used in examples like the one above. Rearrange the non-quadratic equation to find an expression for either x or y. Then substitute this into the quadratic and solve the resulting equation. Substitute the solutions to the quadratic back into the other equation to find the pairs of solutions required. Example Solve this pair of simultaneous equations. x 7 y (7 y) 2 + y 2 25 (49 14y + y 2 ) + y y 2 14y (2y 8)(y 3) 0 y 4 or y 3 When y 4, x When y 3, x Solutions are (3,4) or (4,3) x + y 7, x 2 + y 2 25 Rearrange to give x in terms of y. Substitute x into the quadratic. Multiply out the brackets. Collect like terms and factorise if possible. Solve to find y. Substitute y values into the nonquadratic equation to find pairs of solutions. Now have a go at section 6, questions 2, 4, 6, 8 and 10.

11 Section 7 know the rules for indices be able to simplify expressions that involve indices know the meaning of negative, zero and fractional indices be able to solve equations involving indices. You will have met all these rules before at GCSE, but it is essential you know them off by heart and are confident when using them. Rules for indices Addition rule for multiplication Subtraction rule for division The power-on-power rule The zero index Negative indices x m x n x m+n x m xm n xn (x m ) n x mn x 0 1, for all non-zero values of x. x n 1 x n Fractional indices x 1 n n x

12 Examples Evaluate (a) (b) ( ) 3 2 (C) write in the form 2 n (d) Solve 4 k+1 8 (a) Write as a fraction. Calculate the cube root of (b) ( 25 3 ) 2 ( 25 3 ) ( 5 7 ) Remember whole fraction is to the power of 3 2 Root then cube. (c) (2 2 ) 20 (2 3 ) (d) 4 k+1 8 (2 2 ) k+1 (2) 3 2 2k k k 1 2 Make the base of both the same. Use the power on power rule. Use the addition rule. Make bases the same. Use the power on power rule. Powers are equal. Solve to find k. Now have a go at section 7, questions 1, 2, 3, 4, 8, 9, 11 and 12.

13 Section 8 Surds be able to simplify expressions involving square roots. Be able to leave answers to questions in surd form A number written exactly using square roots is called a surd, e.g. 5 and Rules for surds Multiplication a b a b Examples Division a b a b (a) Express 28 in the form a b, (b) Work out and simplify (a) (b) Note Rationalising the denominator Simplified surds should never have a square root in the denominator. Simplifying the expression to remove the square root from the denominator is called rationalising. Examples (a) Rationalise 1 2 7, (b) Rationalise the denominator of (a) (b) (3 5) (3+ 5)(3 5) Note if the denominator contains an integer and a surd, then change the sign between them before multiplying top and bottom by them. (3 + 5)(3 5) Now have a go at section 8, questions 1, 2, 3, 4, 5, 6, 9, 12, 15, 18, 22, 27, 30, 31, 32.

14 Section 9 Solving quadratics by factorising. be able to solve quadratics by factorising. The simplest way to solve a quadratic is to factorise it if possible. Once it is factorised then use the fact that if the two factors multiply to give 0 then one of the factors must be zero. Example Solve x (x 1)(x + 4) x 2 + 3x 4 0 (x 1)(x + 4) 0 Either (x 1) 0 or (x + 4) 0 So x 1 or x 4 Factorise the quadratic. Make each factor equal to 0 and solve. Now have a go at section 9, questions 1, 4, 7, 9 and 10.

15 Section 10 Solving quadratics by completing the square. be able to solve quadratics by completing the square. Sometimes you need to solve a quadratic that does not factorise, one way of doing this is to complete the square. This method is also useful when you are asked to sketch the graph of a quadratic because it makes identifying the minimum/maximum point (vertex) and the line of symmetry easier. Example Solve x 2 6x by completing the square. x 2 6x Halve the coefficient of x, (-6) 2. Put it into a squared bracket with x, i.e. (x 3) Multiply this bracket out, i.e. x 2 6x Work out what you need to add/subtract to make this equal to the original equation, (-2) (x 2 6x + 9) 2 0 (x 3) (x 3) Put steps 2 and 4 together to give completed square form and solve the resulting equation. x 3 ± 2 x 3 ± 2 Now have a go at all of section 10.

16 Section 11 Solving quadratics using the formula. be able to solve quadratics by using the formula. The factorising method works well if the solutions are integers (whole numbers) or simple fractions. However, if there are no solutions, the solutions involve square roots or the coefficients are horrible then the quadratic formula should be used. The roots of the equation ax 2 + bx + c 0 are x b± b2 4ac 2a Example Use the quadratic formula to find the exact solutions to 2x 2 + 4x (a) 2x 2 + 4x Note x 4 ± ± ± ± The solutions are 1 ± or 1 ± Substitute a 1, b 4 and c1 into the formula. Leave the answer in surd form because this is exact, rounded decimals are not. Also be aware that in Core 1 you will not have a calculator. 1. A level questions will often ask for exact answers, this means if it contains a square root or pi, you should not turn it into a decimal. If it doesn t state otherwise it is ok to give decimal answers, work to 3 significant figures unless told differently. 2. You will not be given the formula you need to learn it. Now have a go at section 11, questions 1, 2, 3, 4 and 5.