STRAND: ALGEBRA Unit 2 Solving Quadratic Equations

Transcription

1 CMM Suject Support Strand: ALGEBRA Unit Solving Quadratic Equations: Tet STRAND: ALGEBRA Unit Solving Quadratic Equations TEXT Contents Section. Factorisation. Using the Formula. Completing the Square

2 CMM Suject Support Strand: ALGEBRA Unit Solving Quadratic Equations: Tet Solving Quadratic Equations. Factorisation Equations of the form a + + c = 0 are called quadratic equations. Many can e solved using factorisation. If a quadratic equation can e written as ( a) ( ) = 0 then the equation will e satisfied if either racket is equal to zero. That is, ( a) = 0 or So there would e two possile solutions, Worked Eample Solve = 0. ( ) = 0 = a and =. Factorising gives So therefore ( + 5) ( + ) = = 0 or + = 0 = 5 or = Worked Eample Solve = 0. Factorising gives So therefore ( ) ( + 7) = 0 = 0 or + 7 = 0 = or = 7 Worked Eample Solve = 0.

3 . CMM Suject Support Strand: ALGEBRA Unit Solving Quadratic Equations: Tet Factorising gives So therefore Worked Eample 4 Solve Factorising gives So therefore ( ) = 0 = 0 or = 0 = 0 or = 4 8 = 0 ( )( + ) = = 0 or + 9 = 0 = 9 or = 9 = 4 = 4 Worked Eample 5 Solve = 0. Factorising gives So therefore ( ) ( ) = 0 = 0 or = 0 = or = This type of solution is often called a repeated solution and results from solving a perfect square, that is 0 ( ) = Most of these eamples have had two solutions, ut the last eample had only one solution.

4 . CMM Suject Support Strand: ALGEBRA Unit Solving Quadratic Equations: Tet The graphs elow show y = and y = y y 5 The curve crosses the -ais at The curve touches the -ais at = 5 and =. = These are the solutions of Eercises This is the solution of = = 0. Solve the following quadratic equations. (a) + = 0 () 5 = 0 (c) + 4 = 0 (d) + 6 = 0 (e) 4 = 0 (f) 4 9 = 0 (g) 9 = 0 (h) 49 = 0 (i) 9 64 = 0 (j) = 0 (k) = 0 (l) 8 = 0 (m) + 8 = 0 (n) + 0 = 0 (o) = 0 (p) = 0 (q) + 5 = 0 (r) = 0 (s) 4 + = 0 (t) + 5 = 0 (u) = 0. The equations of a numer of curves are given elow. Find where each curve crosses the -ais and use this to draw a sketch of the curve. (a) y = () y = 4 (c) y = (d) y = +. Use the difference of two squares result to solve the following equations. (a) 4 6 = 0 () 4 65 = 0 4. Find the lengths of each side of the following rectangles. (a) () Area = Area = + + 4

5 . CMM Suject Support Strand: ALGEBRA Unit Solving Quadratic Equations: Tet (c) (d) Area = Area = The height of a all thrown straight up from the ground into the air at time, t, is given y h = 8t 0t Find the time it takes for the all to go up and fall ack to ground level. 6. The diagram represents a shed. The volume of the shed is given y the formula G V = LW E + R ( ) E R (a) Make L the suject of the formula, giving your answer as simply as possile. W L The surface area, A, of the shed, is given y the formula ( ) A = GL + EL + W E + R where V = 500, A = 00, E = 6 and G = 4. () By sustituting these values into the equations for V and A show that L satisfies the equation L 5L + 50 = 0 Make the steps in your working clear. (c) Solve the equation L 5L + 50 = 0.. Using the Formula The formula given elow is particularly useful for quadratics which cannot e factorised. To prove this important result requires some quite comple analysis, using a technique called completing the square, which is the suject of Section.. Theorem The solutions of the quadratic equation are given y a + + c = 0 ac = ± 4 a 4

6 . CMM Suject Support Strand: ALGEBRA Unit Solving Quadratic Equations: Tet Proof The equation a + + c = 0 is first divided y the non-zero constant, a, giving Note that a c + + = 0 a + = + + a a a = a a a (epanding) = + + a a = + + a a (adding like terms) (simplifying) The first two terms are identical to the first two terms in our equation, so you can re-write the equation as + c a a + a = 0 + = a a c a = 4a c a i.e. + a = 4ac 4a Taking the square root of oth sides of the equation gives + = ± a 4ac 4a Hence as required. or = ± 4 ac a = ± a 4ac a ac = ± 4 a 5

7 . CMM Suject Support Strand: ALGEBRA Unit Solving Quadratic Equations: Tet Worked Eample Solve = 0 giving the solution correct to decimal places. Here a =, = 6 and c = 8. These values can e sustituted into to give ac = ± 4 a ( ) = 6 ± = 6 ± 68 = or 6 68 =. or 7. (to d.p.) Worked Eample Solve the quadratic equation = 0. Here a = 4, = and c = 9. Sustituting the values into gives = ac = ± 4 a ( ) ( ) ± = = ± ± 0 8 = 8 = 5. = 6

8 . CMM Suject Support Strand: ALGEBRA Unit Solving Quadratic Equations: Tet Worked Eample Solve the quadratic equation = 0 Here a =, = and c = 5. Sustituting the values into the formula gives ( ) = ± 4 5 = ± 0 = ± 9 As it is not possile to find 9, this equation has no solutions. These three eamples illustrate that a quadratic equation can have, or 0 solutions. The graphs elow illustrate these graphically and show how the numer of solutions depends on the sign of ( 4ac) which is part of the quadratic formula. y y y Two solutions One solution No solutions 4ac > 0 4ac = 0 4ac < 0 (Worked Eample ) (Worked Eample ) (Worked Eample ) Eercises. Use the quadratic equation formula to find the solutions, where they eist, of each of the following equations. Give answers to decimal places. (a) = 0 () + 0 = 0 (c) 9 6 = 0 (d) 5 7 = 0 (e) + 8 = 0 (f) = 0 (g) = 0 (h) 4 = 0 (i) + 0 = 0 (j) + 8 = 0 (k) + 6 = 0 (l) 8 + = 0 (m) 4 5 = 0 (n) = 0 (o) 6 5 = 0 7

9 . CMM Suject Support Strand: ALGEBRA Unit Solving Quadratic Equations: Tet. A ticket printing and cutting machine cuts rectangular cards which are cm longer than they are wide. (a) If is the width of a ticket, find an epression for the area of the ticket. () Find the size of a ticket with an area of 0 cm.. A window manufacturer makes a range of windows for which the height is 0.5 m greater than the width. Find the width and height of a window with an area of m. 4. The height of a stone launched from a catapult is given y h = 0t 9. 8t where t is the time after the moment of launching. (a) () (c) (d) Find when the stone hits the ground. For how long is the stone more than 5 m aove the ground? Is the stone ever more than m aove ground level? If m is the maimum height of the stone, write down a quadratic equation which involves m. Eplain why this equation has only one solution and use this fact to find the value of m, to decimal places. 5. The equation elow is used to find the maimum amount,, which a ungee cord stretches during a ungee jump: mg + mgl k = 0, where m = mass of ungee jumper l = length of rope when not stretched (0 m) k = stiffness constant 0 (a) () g = acceleration due to gravity 0 ( Nm ) ( ms ) Find the maimum amount that the cord stretches for a ungee jumper of mass 60 kg. How much more would the cord stretch for a person of mass 70 kg? 6. Solve the equation = 5 + 7, giving your answers correct to significant figures. 8

10 CMM Suject Support Strand: ALGEBRA Unit Solving Quadratic Equations: Tet. Completing the Square Completing the square is a technique which can e used to solve quadratic equations that do not factorise. It can also e useful when finding the minimum or maimum value of a quadratic. A general quadratic a + + c is written in the form a( + p) + q when completing the square. You need to find the constants p and q so that the two epressions are identical. Worked Eample Complete the square for First consider the ( ) + 0. These terms can e otained y epanding + 5. ( ) = = ( + ) But so ( ) + ( 5) Therefore = Worked Eample Complete the square for To otain ( ) + 6 requires epanding +. = + ( ) = + + But ( ) so + 6 = + 9 ( ) Therefore = Note ( ) = + 7 When completing the square for + + c, the result is + + c = + and for a 0, a + + c = a + c c a 4a 9

11 . CMM Suject Support Strand: ALGEBRA Unit Solving Quadratic Equations: Tet Worked Eample Complete the square for As a first step, the quadratic can e rearranged as shown elow. ( ) = + 7 Then note that + = + ( ) + = + ( ) [( ) ] + ( ) + so Worked Eample 4 (a) Complete the square for y = 8 +. () Find the minimum value of y. (c) Sketch the graph of y = 8 +. (a) First rearrange the quadratic as shown. Then = + 7 ( ) + = + 4 ( ) = 4 ( ) to give [( ) ] + 4 can e written as 4 ( ) ( ) + = 4 4 ( ) + = 8 ( ) = 6 () As y = 6, the minimum possile value of y is 6, which is otained when = 0 or =. (c) Before sketching the graph, it is also useful to find where the curve crosses the -ais, that is when y = 0. To do this, solve ( ) 0 = 6 ( ) = 6 ( ) = = ± = ± So the curve crosses the -ais at + and, and has a minimum at (, 6). 0

12 . CMM Suject Support Strand: ALGEBRA Unit Solving Quadratic Equations: Tet This is shown in the graph opposite. y Worked Eample 5 6 (, 6) ( ) + (a) Epress + + in the form a + p q numers. where a, p and q are real ( ) = + + () Hence, determine for f (i) (ii) the minimum value for f ( ) the equation of the ais of symmetry. ( ) + (a) + + = a + p q Equating coefficients: ( ) + = a + p + p q ( ) = a + ap + ap + q [ ] = a a = [ ] = ap p = = a [ c t ] = ap + q = + q = + q Thus q = = + + = + +

13 . CMM Suject Support Strand: ALGEBRA Unit Solving Quadratic Equations: Tet () (i) Minimum value of y = + + =, and the value is y =. will occur when + = 0 ; that is, (ii) y = is the equation of the ais of symmetry. Eercises. Complete the square for each of the epressions elow. (a) () + 6 (c) + 0 (d) 8 + (e) + + (f) (g) + (h) 5 + (i) + 4. Use the completing the square method to solve each of the following equations. (a) 4 + = 0 () 6 4 = 0 (c) = 0 (d) = 0 (e) + = 0 (f) + 4 = 0 (g) = 0 (h) + 5 = 0 (i) = 0. Complete the square for each of the following epressions. (a) + 8 () + 0 (c) + + (d) + 6 (e) (f) (g) + 4 (h) (i) +

14 . CMM Suject Support Strand: ALGEBRA Unit Solving Quadratic Equations: Tet 4. Solve each of the following equations y completing the square. (a) = 0 () + 6 = 0 (c) + 8 = 0 (d) 4 + = 0 (e) + 6 = 0 (f) = 0 5. Sketch the graph of each equation elow, showing its minimum or maimum point and where it crosses the -ais. (a) y = () y = (c) y = (d) y = (e) y = 4 + (f) y = 6. The height of a all thrown into the air is given y h = + 0t 0t Find the maimum height reached y the all. 7. (a) By writing the quadratic epression ( ) in the form + a, find a and and hence find the minimum value of the epression. () Solve the equation 4 + = 0 giving your answers correct to decimal places. 8. (a) Factorise the epression +. ( ) () Epress + in the form + a, where a and are whole numers. (c) Sketch the curve with equation y = +. ( ) = 9. (a) Epress the function f 4 in the form ( ) = ( + ) + f a h k. Hence, or otherwise, determine () the values of at which the graph cuts the -ais. (c) the interval for which f( ) 0

15 . CMM Suject Support Strand: ALGEBRA Unit Solving Quadratic Equations: Tet (d) (e) the minimum value of f( ) the value of at which f( ) is a minimum. 0. (a) If 4y + y + is a perfect square, calculate the value of. () By the method of completing the square, solve the equation 5y = 8y. Give your answers to significant figures. Information The word 'quadratic' comes from the Latin word 'quadratum', which means 'a squared figure'. 4