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## Algebra

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**EXPANDING**- Does 6 × (3 + 5) = 6 × 3 + 6 × 5 ? YES 6 × 8 = 18 + 30 48 = 48 - The removal of the brackets is known as the distributive law and can also be applied to algebraic exressions - When expanding, simply multiply each term inside the bracket by the term directly in front e.g. Expand a) 6(x + y) = 6 × x + 6 × y b) -4(x – y) = -4 × x - -4 × y = 6x + 6y = -4x + 4y c) -4(x – 6) = -4 × x - -4 × 6 d) 7(3x – 2) = 7 × 3x - 7 × 2 = -4x + 24 = 21x - 14 e) x(2x + 3y) = x × 2x 1 1 + x × 3y f) -3x(2x – 5) = -3x × 2x - -3x × 5 = 2x2 + 3xy = -6x2 + 15x Don’t forget to watch for sign changes!**- If there is more than one set of brackets, expand them all**then collect any like terms. e.g. Expand and simplify a) 2(4x + y) + 8(3x – 2y) = 2 × 4x + 2 × y + 8 × 3x - 8 × 2y = 8x + 2y + 24x - 16y = 32x - 14y b) -3(2a – 3b) – 4(5a + b) = -3 × 2a - -3 × 3b - 4 × 5a + -4 × 1b = -6a + 9b - 20a - 4b = -26a + 5b**EXPANDING TWO BRACKETS**- To expand two brackets, we must multiply each term in one bracket by each in the second Remember integer laws when multiplying e.g. Expand and simplify a) (x + 5)(x + 2) = x2 + 2x + 5x + 10 b) (x - 3)(x + 4) = x2 + 4x - 3x - 12 = x2 + 7x + 10 = x2 + 1x – 12 To simplify, combine like terms c) (x - 1)(x - 3) = x2 - 3x - 1x + 3 c) (2x + 1)(3x - 4) = 6x2 - 8x + 3x - 4 = x2 – 4x + 3 = 6x2 – 5x – 4**PERFECT SQUARES**- When both brackets are exactly the same To simplify, combine like terms e.g. Expand and simplify Watch sign change when multiplying a) (x + 8)2 = (x + 8)(x + 8) b) (x - 4)2 = (x – 4)(x – 4) = x2 + 8x + 8x + 64 = x2 - 4x - 4x + 16 = x2 + 16x + 64 = x2 - 8x + 16 Write out brackets twice BEFORE expanding c) (3x - 2)2 = (3x – 2)(3x – 2) = 9x2 - 6x - 6x + 4 = 9x2 – 12 x + 4**DIFFERENCE OF TWO SQUARES**- When both brackets are the same except for signs (i.e. – and +) e.g. Expand and simplify a) (x – 3)(x + 3) = x2 + 3x - 3x - 9 b) (x – 6)(x + 6) = x2 + 6x - 6x - 36 = x2 – 9 = x2 – 36 Like terms cancel each other out c) (2x – 5)(2x + 5) = 4x2 + 10x - 10x - 25 = 4x2 – 25**MORE THAN TWO BRACKETS**- Expand two brackets at a time e.g. Expand and simplify a) (x + 2)(x - 3)(x + 5) = (x2 - 3x + 2x - 6) (x + 5) = (x2 – x – 6) (x + 5) = x3 + 5x2 - x2 - 5x - 6x - 30 = x3 + 4x2 – 11x - 30 b) (x - 2)3 = (x – 2)(x – 2)(x – 2) = (x2 - 2x - 2x + 4) (x - 2) = (x2 – 4x + 4) (x – 2) = x3 - 2x2 - 4x2 + 8x + 4x - 8 = x3 - 6x2 + 12x - 8**FACTORISING**- Factorising is the reverse of expanding - To factorise: 1) Look for a common factor to put outside the brackets 2) Inside brackets place numbers/letters needed to make up original terms You should always check your answer by expanding it e.g. Factorise a) 2x + 2y = 2( ) x + y b) 2a + 4b – 6c = 2( ) a + 2b - 3c - Always look for the highest common factor e.g. Factorise a) 6x - 15 = 3( ) 2x - 5 b) 30a + 20 = 10( ) 3a + 2 - Sometimes a ‘1’ will need to be left in the brackets e.g. Factorise a) 6x + 3 = 3( ) 2x + 1 b) 20b - 10 = 10( ) 2b - 1**- Letters can also be common factors**e.g. Factorise a) cd - ce = c( ) d - e b) xyz + 2xy – 3yz = y( ) xz + 2x - 3z c) 4ad – 8a = 4 ( ) a d - 2 - Powers greater than 1 can also be common factors e.g. Factorise a) 5a2 – 7a5 = a2( ) 5 - 7a3 b) 4b2 + 6b3 = 2b2( ) 2 + 3b**Factorising by grouping**- When two groups within an expression have their own common factor e.g. Factorise As both 2c and 3d are being multiplied by (a + b) we place them in separate brackets 2ac + 2bc + 3ad + 3bd = 2c( ) a + b + 3d( ) a + b = (2c + 3d)(a + b)**FACTORISING QUADRATICS**The general equation for a quadratic is ax2 + bx + c When a= 1 - You need to find two numbers that multiply to give c and add to give b e.g. Factorise To check answer, expand and see if you end up with the original question! a) x2 + 11x + 24 = (x + 3)(x + 8) 1, 24 List pairs of numbers that multiply to give 24 (c) Check which pair adds to give 11 (b) Place numbers into brackets with x 2, 12 3, 8 4, 6 b) x2 + 7x + 6 = (x + 1)(x + 6) 1, 6 List pairs of numbers that multiply to give 6 (c) Check which pair adds to give 7 (b) Place numbers into brackets with x 2, 3**- Expressions can also contain negatives**e.g. Factorise a) x2 + x – 12 = (x - 3)(x + 4) b) x2 – 6x – 16 = (x + 2)(x - 8) - 1, 12 1, 16 - As the end number (c)is -12, one of the pair must negative. As the end number (c)is -16, one of the pair must negative. - 2, 6 2, 8 - - 3, 4 4, 4 - Check which pair now adds to give b Make the biggest number of the pair the same sign as b Check which pair now adds to give b Make the biggest number of the pair the same sign as b c) x2 – 9x + 20 = (x - 4)(x - 5) - 1, 20 - As the end number (c) is +20, but b is – 9, both numbers must be negative - 2, 10 - - 4, 5 - Check which pair now adds to give b**SPECIAL CASES**1. Perfect Squares e.g. Factorise a) x2 + 10x + 25 = (x + 5)(x + 5) b) 4x2 - 16x + 16 = 4(x2 – 4x + 4) 1, 25 = (x + 5)2 = 4(x– 2)(x – 2) - 1, 4 - 5, 5 = 4(x – 2)2 - 2, 2 - 2. Difference of Two Squares e.g. Factorise a) x2 + 0x - 25 = (x - 5)(x + 5) b) 9x2 – 100 = (3x )(3x ) - 10 + 10 -5, 5 c) 2x2 – 72 = 2(x2 – 36) Add in a zero x term and factorise = 2(x )(x ) - 6 + 6 OR: factorise by using A2 – B2 = (A – B)(A + B)**TWO STAGE FACTORISING**When a≠ 1 1. Common factor - Always try to look for a common factor first. e.g. Factorise a) 2x2 + 12x + 16 = 2( ) x2 + 6x + 8 b) 3x2 – 6x – 9 = 3( ) x2 – 2x – 3 1, 8 = 2(x + 2)(x + 4) 1, 3 - = 3(x + 1)(x – 3) 2, 4 c) 3x2 + 24x = 3x( ) x + 8 d) 4x2 – 36 = 4( ) x2 – 9 = 4(x )(x ) - 3 + 3**2. No common factor (HARD)**- Use the following technique e.g. Factorise a) 3x2 – 10x – 8 = 3x2 + 2x – 12x – 8 = x( ) 3x + 2 -4( ) 3x + 2 = (x – 4)(3x + 2) 1, 24 - Multiply constants of first and last terms Find two nos.that multiply to -24 but add to -10 Replace 10x with the two new terms Factorise 2 terms at a time. 2, 12 - 3, 8 - 4, 6 - 3 × - 8 = -24 Write in two brackets b) 2x2 + 7x + 3 = 2x2 + x + 6x + 3 = x( ) 2x + 1 + 3( ) 2x + 1 = (x + 3)(2x + 1) 2 × 3 = 6 1, 6 2, 3**RATIONAL EXPRESSIONS**- Look to factorise the numerator and/or denominator and then remove the common factor by dividing through. Cancel out common factors e.g. x + 3 = 1 x + 3 e.g. Simplify a) x2 + 7x + 12 x + 3 = (x + 3)(x + 4) x + 3 = (x + 4) As the numerator cancels out, you must leave a ‘1’ on top to signal this b) x + 7 . 49 – x2 = x + 7 . (7 – x)(7 + x) = 1 . (7 – x) - Beware of disguised sign changes i.e. (x – 7) cancels with (7 – x) leaving -1 e.g. Simplify x2 – 5x + 6 8 – 2x – x2 = (x – 3)(x – 2) (x + 4)(2 – x) = -1 (x – 3) (x + 4) Better written as: -(x – 3) (x + 4)**POWER RULES**1. Multiplication - Does x2 × x3 = x × x × x × x × x ? YES - Therefore x2 × x3 = x5 - How do you get 2 3 = 5 ? + - When multiplying index (power) expressions with the same letter, ADD the powers. No number = 1 i.e. p = 1p1 e.g. Simplify a) p10 × p2 = p(10 + 2) b) a3 × a2 × a = a(3+ 2 + 1) 1 = p12 = a6 - Remember to multiply any numbers in front of the variables first e.g. Simplify a) 2x3 × 3x4 = 2 × 3 x(3+ 4) b) 2a2 × 3a × 5a4 1 = 6 x7 = 2 × 3 × 5 a(2 + 1 + 4) = 30 a7**2. Division**- Does 6 = 1 ? 6 YES - Therefore x = 1 x - Does x5 = x × x × x × x × x? x3 x × x × x YES = x × x × 1 × 1 × 1 - Therefore x5 = x2 x3 - How do you get 53 = 2 ? - - When dividing index (power) expressions with the same letter, SUBTRACT the powers. e.g. Simplify a) p5 ÷ p = p(5 - 1) b) x7 x4 = x(7 - 4) 1 = p4 = x3 - Remember to divide any numbers in front of the variables first e.g. Simplify a) 12x5 ÷ 6x4 = 12 ÷ 6 x(5- 4) b) 5a7 15a2 ÷ 5 ÷ 5 = 1 5 a(7- 2) = 2 x = 1 5 a5 or a5 5 If the power remaining is 1, it can be left out of the answer**3. Powers of powers**- Does (x2)3 = x2 × x2 × x2 ? YES - Does x2 × x2 × x2 = x6 ? YES - Therefore (x2)3 = x6 - How do you get 2 3 = 6 ? × - When taking a power of an index expression to a power, MULTIPLY the powers e.g. Simplify a) (c4)6 = c(4 × 6) b) (a3)3 = a(3× 3) = c24 = a9 - If there is a number in front, it must be raised to the power, not multiplied e.g. Simplify a) (3d2)3 = 33 × d(2 × 3) b) (2a3)4 = 24 × a(3 × 4) = 27 d6 = 16 a12 - If there is more than one term in the brackets, raise all to the power e.g. Simplify b) (4b2c5)2 = 42 b(2 × 2) c(5 × 2) a) (x3y z4)3 = x(3 × 3) y(1 × 3) z(4 × 3) 1 = x9 y3 z12 = 16 b4 c10**NEGATIVE POWERS**In General: x-n = 1 xn e.g. Write with a positive index a) x-4 = 1 x4 b) 3x-1 = 3 x e.g. Evaluate a) (3)-3 = 1 33 b) (0.5)-2 = 4 Therefore 1-2 = 2 22 1 = 1 27 = 4 e.g. Write 2-2 with a positive index and then evaluate 5 52 2 = 25 4**NEGATIVE POWER LAWS**- Same laws learned previously apply to negative powers e.g. Simplify the following and write with a positive index a) 3x-2 × 4x-3 = 3 × 4 x-2+ -3 b) (xy2)-4 = x1 × -4 y2 × -4 = 12x-5 = x-4y-8 = 12 x5 = 1 . x4y8 FRACTIONAL INDICES - Express roots of numbers e.g. The power of a ½ is the same as square root. In General: Index form Surd form**e.g. Write x1/4 in surd form**e.g. Evaluate 641/3 e.g. Write in index form = x1/5 e.g. Simplify (64x6)1/2 = √64 x6 x 1/2 = 8x3 COMBINING POWERS & ROOTS In General: e.g. Evaluate 642/3 e.g. Write in index form**FRACTIONAL & NEGATIVE POWERS**e.g. Write x-2/3 in surd form e.g. Evaluate**ALGEBRAIC FRACTIONS**1. Simplifying - Always take out any common factors. a) 3ab 9bd = 1a 3d b) 12x2y3z 27xz2 = 4xy3 9z = a 3d 2. Multiplying Fractions - Multiply top and bottom terms separately then simplify. e.g. Simplify: a) 4a × 3e 5e 2a = 4a × 3e b) 5x2y × 6y2 4x5x3y = 5x2y × 6y2 5e × 2a 4x × 5 x3y = 30x2y3 20x4y = 12ae 10ae = 3y2 2x2 = 6 5**2. Dividing Fractions**- Multiply the first fraction by the reciprocal of the second, then simplify Note: b is the reciprocal of 2 2 b e.g. Simplify: a) 2a ÷ a2 5 3 = 2a 5 × 3 a2 b) 6r ÷ 3r2 5 = 6r 5 × 1 3r2 = 2a × 3 = 6r × 1 5 × a2 5 × 3r2 = 6a 5a2 = 6r 15r2 = 6 5a = 2 5r**4. Adding/Subtracting Fractions**a) With the same denominator: - Add/subtract the numerators and leave the denominator unchanged. Simplify if possible. e.g. Simplify: a) 3x + 3x 10 10 = 3x + 3x 10 b) 6a - b 5 5 = 6a - b 5 = 6x 10 ÷ 2 ÷ 2 = 3x 5 b) With different denominators: - Multiply denominators to find a common term. - Cross multiply to find equivalent numerators. - Add/subtract fractions then simplify. e.g. Simplify: b) 3 – 2 ab b×3 = a×b - a×2 a) a + 2a 2 3 = 2×3 3×a + 2×2a = 3a + 4a 6 = 3b – 2a ab = 7a 6**Harder examples:**a) 3x + 2 + 2 4x 6x 6x(3x + 2) = 4x×6x + 4x×2 b) 5 . – 2 . x – 1 x + 6 = (x – 1)(x + 6) 5(x + 6) - 2(x – 1) = 18x2 + 12x + 8x 24x2 = 5x+ 30 - 2x + 2 x2 + 5x - 6 = 18x2 + 20x 24x2 = 3x+ 32 . x2 + 5x - 6 = 2x(9x + 10) 2x×12x = 9x + 10 12x Always look for common factors to simplify answer!**LOGARITHMS**1. Formal definition: If bp = q then logb(q) = p Where b = base, p = logarithm and q = number For simple questions use guess and check. e.g. a) Write 25 = 32 in log form log2(32) = 5 b) Write log5(125) = 3 in index form 53 = 125 As less than 1, power is negative. e.g. Calculate a) log4(1024) 4x = 125 b) log10(0.01) 10x = 0.01 x = 5 x = -2 e.g. Solve a) log6(x) = 3 63 = x b) logx(243) = 5 x5 = 243 x = 216 x = 5√243 x = 3**2. Properties of Logarithms:**- log (ab) = log(a) + log (b) - log (a/b) = log(a) – log (b) - log (an) = nlog(a) e.g. Simplify a) log(4) + log(6) = log(4 × 6) b) log(27) – log(9) = log(27 ÷ 9) = log(24) = log(3) c) Write 4log(2) as the log of a single number = log(24) = log(16) Remember your BEDMAS! e.g. Simplify a) 3log(2) + log(4) – log(8) • log(27) • log(3) = log(33) log(3) = log(23) + log(4) – log(8) = 3log(3) log(3) = log(8) + log(4) – log(8) = log(32) – log(8) = 3 = log(4)**Use the properties of logs to move the unknown**3. Index Equations: - When x is the ‘power’ and it isn’t a simple whole number answer - Solved by taking the logs of both sides Simply use log button to find answer e.g. Solve a) 3x = 21 b) 2x = 30 log3x = log21 log2x = log30 xlog3 = log21 xlog2 = log30 x = log21 log3 x = log30 log2 x = 2.77 (2 d.p.) x = 4.91 (2 d.p.) c) 134x- 5 = 6 d) Evaluate log4(128) log134x- 5 = log6 4x = 128 (4x – 5)log13 = log6 log4x = log128 4x – 5 = log6 log13 xlog4 = log128 x = log128 log4 4x – 5 = 0.699 4x = 5.699 x = 3.5 x = 1.42 (2 d.p.) Make sure to use full, not rounded answers throughout question**SOLVING LINEAR EQUATIONS**- Remember that addition/subtraction undo each other as do multiplication/division e.g. Solve a) 3x – 1 = 3 – x b) 3(x – 1) = 5x + 17 +x +x 3x – 3 = 5x + 17 4x – 1 = 3 -5x -5x +1 +1 -2x – 3 = 17 4x = 4 +3 +3 ÷4 ÷4 -2x = 20 x = 1 ÷-2 ÷-2 x = -10 - Not every linear equation has a unique solution 1) Some have an infinite number of solutions (Solution set is R, the set of real numbers) e.g. Solve 6x + 5 = 2(3x + 5) – 5 6x + 5 = 6x + 10 – 5 As the answer is considered a ‘true statement’, x can be any value possible 6x + 5 = 6x + 5 -6x -6x 0 + 5 = 5 -5 -5 0 = 0**2) Some have no solutions (Solution set is an empty set)**e.g. Solve 2x – 3 = 2x + 7 As the answer is NOT considered a ‘true statement’, there are NO possible values for x -2x -2x 0 – 3 = 7 +3 +3 0 = 10 Equations in Fraction Form - If equations are fraction only, cross multiply, then solve e.g. Solve a) x = 9 4 2 b) 3x - 1 = x + 3 5 2 2x = 36 2(3x - 1) = 5(x + 3) ÷2 ÷2 x = 18 6x - 2 = 5x + 15 -5x -5x x - 2= 15 +2 +2 x = 17**- For two or more fractions, find a common denominator,**multiply it by each term, then solve e.g. Solve 4x - 2x = 10 5 3 ×15 ×15 ×15 5 × 3 = 15 60x 5 - 30x 3 = 150 Simplify terms by dividing numerator by denominator 12x - 10x = 150 2x = 150 ÷2 ÷2 x = 75 e.g. Solve 5x - (x + 1) = 2x 6 4 ×24 ×24 ×24 120x 6 - (24x + 24) 4 = 48x 20x - 6x – 6 = 48x 14x – 6 = 48x -48x -48x -34x – 6 = 0 + 6 + 6 -34x = 6 ÷ -34 ÷ -34 x = -6 34**SOLVING INEQUATIONS**- Inequations contain one of four inequality signs: < > ≤ ≥ - To solve follow the same rules as when solving equations - Except: Reverse the direction of the sign when dividing by a negative e.g. Solve a) 3x + 8 >24 b) -2x -5 ≤ 13 -8 -8 +5 +5 3x >16 -2x ≤ 18 As answer not a whole number, leave as a fraction ÷3 ÷3 ÷-2 ÷-2 Sign reverses as dividing by a negative x >16 3 x ≥ -9 c) 4x – 3(x + 6) ≥ 5x – 22 4x – 3x – 18 ≥ 5x – 22 x – 18 ≥ 5x – 22 +18 +18 x ≥ 5x – 4 -5x -5x Sign reverses as dividing by a negative -4x ≥ – 4 ÷-4 ÷-4 x ≤ 1**WRITING AND SOLVING**e.g. Write an equation for the following information and solve a) A rectangular pool has a length 5m longer than its width. The perimeter of the pool is 58m. Find its width x + 5 x + 5 + x + x + 5 + x = 58 4x + 10 = 58 Draw a diagram -10 -10 x x 4x = 48 Let x = width ÷4 ÷4 x + 5 x = 12 Therefore width is 12 m b) I think of a number and multiply it by 7. The result is the same as if I multiply this number by 4 and add 15. What is this number? Let n = a number 7 n = n 4 + 15 -4n -4n 3n = 15 ÷3 ÷3 n = 5 Therefore the number is 5**REARRANGING FORMULA**- Involves rearranging the formula in order to isolate the new ‘subject’ - Same rules as for solving are used Remember: When rearranging or changing the subject you are NOT finding a numerical answer 1. Linear a) Make x the subject of y = 6x - 2 +2 +2 y + 2 = 6x ÷6 ÷6 All terms on the left must be divided by 6 y + 2 = x 6 b) Make R the subject of IR = V Treat letters the same as numbers! ÷I ÷I R =V I**2. Rational Expressions**- Steps: Cross multiply All x terms to the left, others to the right Factorise, then divide e.g. Make x the subject of: 3y(2x – 1) = 5 y(x + 3) = 5x – 2 6xy – 3y = 5 xy + 3y = 5x – 2 6xy = 5 + 3y xy – 5x = – 2 – 3y x(y – 5) = – 2 – 3y 3. Square Roots - Squaring undoes square roots e.g. Make x the subject of: y2 = 3x – 1 y2 + 1 = 3x (3y + 4)2 = 2x + 5 9y2 + 24y + 16 = 2x + 5 9y2 + 24y + 11 = 2x Make sure the square root is on it’s own**4. Squares**a) Perfect Squares - If there’s a single perfect square term in x take the square root e.g. Make x the subject of: Don’t forget the +/- sign! b) Non Perfect Squares (harder!) - If there are single terms of x we must ‘complete the square’ e.g. Write x2 + 10x – 3 in perfect square form: - Steps: -Take half the co-efficient of x 10 ÷ 2 = 5 -Square it 52 = 25 -Add square to the expression and factorise perfect square = (x2 + 10x + 25)– 3 = (x + 5)2– 3 = (x + 5)2– 3 – 25 -Subtract square to keep original expression = (x + 5)2– 28**e.g. Make x the subject of y = x2 – 8x + 5**First turn into a perfect square -8 ÷ 2 = -4 (-4)2 = 16 y = (x2 – 8x + 16)+ 5 Then rearrange equation as before y = (x – 4)2+ 5 y = (x – 4)2+ 5 – 16 y = (x – 4)2– 11 y + 11 = (x – 4)2**SOLVING QUADRATICS**To solve use the following steps: 1. Move all of the terms to one side, leaving zero on the other 2. Factorise the equation 3. Set each factor to zero and solve. e.g. Solve a) (x + 7)(x – 2) = 0 b) x2 – 5x – 6 = 0 1, 6 - 2, 3 - x + 7 = 0 x – 2 = 0 (x + 1)(x – 6) = 0 -7 -7 +2 +2 x + 1 = 0 x – 6 = 0 x = - 7 x = 2 -1 -1 +6 +6 x = -1 x = 6 c) 3x2 – 21x + 36 = 0 - 1, 12 - d) x2 = 4x + 5 - 2, 6 - -4x -5 -4x -5 1, 5 - 3(x2 – 7x + 12) = 0 - 3, 4 - x2 – 4x – 5 = 0 3(x – 3)(x – 4) = 0 (x + 1)(x – 5) = 0 x – 3 = 0 x – 4 = 0 +3 +3 +4 +4 x + 1 = 0 x – 5 = 0 x = 3 x = 4 -1 -1 +5 +5 x = -1 x = 5**e) 2x2 = 5x**f) x2 = 49 x2 – 49 = 0 2x2 – 5x = 0 (x )(x ) = 0 - 7 + 7 x( ) = 0 2x – 5 x - 7 = 0 x + 7 = 0 x = 0 2x – 5 = 0 +7 +7 - 7 - 7 + 5 + 5 x = 7 x = -7 2x = 5 ÷ 2 ÷ 2 x = 5/2**QUADRATIC EQUATIONS**- Involves writing an equation from the information then solving e.g. The product of two consecutive numbers is 20. What are they? If x = a number, then the next consecutive number is x + 1 x + 5 x(x + 1) = 20 - 1, 20 x2 + x = 20 - 2, 10 -20 -20 A = 150 m2 x - 4, 5 x2 + x – 20 = 0 (x – 4)(x + 5) = 0 x – 4 = 0 x + 5 = 0 +4 +4 -5 -5 x(x + 5) = 150 x = 4 x = -5 x2 + 5x = 150 The numbers are 4, 5 and -5, -4 -150 -150 x2 + 5x – 150 = 0 (x – 10)(x + 15) = 0 e.g. A paddock of area 150 m2 has a length 5 m longer than its width. Find the dimensions of this paddock. x – 10 = 0 x + 15 = 0 +10 +10 -15 -15 x = 10 x = -15 The dimensions are 10 m by 15 m**SOLVING BY COMPLETING THE SQUARE**- When certain equations cannot be factorised - Steps to follow for equations of the form ax2 + bx + c = 0 - Move constant to the other side of the equation - Add (b ÷ 2)2 to both sides - Write left side as a perfect square - Take square root and rearrange e.g. Solve x2 – 8x – 17 = 0 Don’t forget the +/- sign! x2 – 8x = 17 x2 – 8x + (-4)2 = 17 + (-4)2 It is OK to leave answer without getting two decimal results x2 – 8x + 16 = 17 + 16 x2 – 8x + 16 = 33 (x – 4)2 = 33 It is important when completing the square that a= 1 x – 4 = ±√33 x = 4 ± √33**QUADRATIC FORMULA**- An important concept to know, especially when equations do not factorise easily (i.e. factors are not whole numbers) - For the general equation ax2 + bx + c = 0 use: Solutions are also know as ROOTS Substitute into formula Make equation equal to zero e.g. Use the quadratic formula to solve a) x2 + 5x + 1 = 0 b) 4x2 – 7x = -2 4x2 – 7x + 2 = 0 a = 1 b = 5 c = 1 a = 4 b = -7 c = 2 x = -0.209 or -4.791 x = 1.390 or 0.360 It is OK to leave answer without getting two decimal results**NATURE OF ROOTS**- Quadratic expressions have parabolas for graphs - The solutions (roots) of an equation are where the parabola cuts the x axis. - There are 3 possibilities 1. Parabola cuts x axis at 2 points 2. Parabola touches x axis at a single point • Equation has 2 real roots • (solutions) e.g. x = 1 or 4 • Equation has 1 repeated root • (solution) e.g. x = 3**3. Parabola does not cross x axis at all**• Equation has no real roots • (solutions)**DISCRIMINANT OF ROOTS**- Tells us the number of real roots an equation has. - For the general equation ax2 + bx + c = 0 the discriminant is: Note: If b2 – 4ac > 0 and a perfect square (i.e. 9, 16, 25…) there are 2 rational roots. This means both roots can be written as a fraction and the quadratic equation factorises. 1. If b2 – 4ac > 0 there are two real roots 2. If b2 – 4ac = 0 there is one repeated root 3. If b2 – 4ac < 0 there are no real roots e.g. Calculate the discriminant of the following and hence state the number of roots a) x2 + 5x + 1 = 0 b) 3x2 – 2x + 8 = 0 c) 4x2 + 5x + 1 = 0 a = 1 b = 5 c = 1 a = 3 b = -2 c = 8 a = 4 b = 5 c = 1 = 21 = -92 = 9 As b2 – 4ac > 0 there are two real roots As b2 – 4ac < 0 there are no real roots As b2 – 4ac > 0 there are two real roots As b2 – 4ac = 9 there are also two rational roots**PROPERTIES OF DISCRIMINANT**e.g. The equation 3x2 – 4x + q = 0 has two real roots. Find the range of possible values of q. As there are two real roots, b2 – 4ac > 0 a = 3 b = -4 c = q Sign reverses as dividing by a negative**SIMULTANEOUS EQUATIONS**- Are pairs of equations with two unknowns Revision of Linear-Linear Equations using two methods: 1. ELIMINATION METHOD - Line up equations and either add or subtract so one variable disappears e.g. Solve To remove the ‘y’ variable we add as the signs are opposite. To remove the ‘y’ variable we subtract as the signs are the same. • 2x + y = 20 • x – y = 4 b) 2x + y = 7 x + y = 4 + ( ) - ( ) 3x = 24 x = 3 ÷3 ÷3 Now we substitute x-value into either equation to find ‘y’ 2 × 3 + y = 7 Now we substitute x-value into either equation to find ‘y’ x = 8 6 + y = 7 2 × 8 + y = 20 -6 -6 16 + y = 20 y = 1 -16 -16 y = 4 Check values in either equation Check values in either equation 8 – 4 = 4 3 + 1 = 4**- You may need to multiply an equation by a number to be**able to eliminate a variable e.g. Solve b) 4x – 2y = 28 3x + 3y = 12 12x – 6y = 84 12x + 12y = 48 • 2x – y = 0 • x + 2y = 5 × 2 × 1 4x – 2y = 0 x + 2y = 5 × 3 × 4 + ( ) - ( ) Multiply the 1st equation by ‘2’ then add to eliminate the y 5x = 5 Multiply the 1st equation by ‘3’ and the 2nd by ‘4’ then subtract to eliminate the x -18y = 36 ÷5 ÷5 ÷-18 ÷-18 x = 1 y = -2 2 × 1 – y = 0 4x – 2 ×-2 = 28 2 – y = 0 4x + 4 = 28 Now we substitute x-value into either equation to find ‘y’ Now we substitute y-value into either equation to find ‘x’ -2 -2 -4 -4 – y = -2 4x = 24 ÷-1 ÷-1 ÷4 ÷4 y = 2 x = 6 Check values in either equation Check values in either equation 1 + 2 ×2 = 5 3 × 6 + 3 ×-2 = 12 Note that it was possible to eliminate the ‘x’ variable by multiplying second equation by 2 and then subtracting Note that it was possible to eliminate the ‘y’ variable by multiplying the 1st equation by 3 and the 2nd by 2 and then adding