MEI C1 Study Resources Core1 Basic Algebra 2 Quadratics

MEI Core 1 Basic Algebra Section 2: Quadratics Study plan Background As for section 1, you will have met most, if not all, of this work at GCSE. Quadratics is a fundamental topic at AS, which links to a lot of the other topics. You need to master this work thoroughly. Detailed work plan 1. Read through the section on Quadratic factorisation on pages 14 to 18 and spend some time following the examples. This should be mostly revision of GCSE work, although you may not have seen examples like Example 1.27 and 1.28 before. There are further worked examples in the Notes and Examples. 2. You may find the PowerPoint demonstration Factorising quadratics and the Factorising quadratics video useful. 3. Exercise 1D Try the first, a middle and the last part of questions 1 to 5. If you find them difficult, do some other parts of each question until you feel confident. You may download worked solutions to questions 2*, 4* and 5*. 4. For additional practice, use the interactive questions Factorising quadratics. You can also test yourself on factorising using the Flash resource Factorising quadratics. 5. Read the section on Solving quadratic equations on pages 18 to 23, This may well go beyond what you have seen at GCSE and requires careful study. There are further worked examples in the Notes and Examples. Note that the method of completing the square is used to derive the quadratic formula, and it is more efficient to use the formula rather than the completing the square method to solve quadratic equations which do not factorise. 6. Try some of the interactive resources on the website. There are Flash resources on Quadratic equations 1 (factorising, with the coefficient of x² as 1), Quadratic equations 2 (factorising, with the coefficient of x² greater than 1), the discriminant, Quadratic equations 3 (using the formula) and Quadratic equations 4 (completing the square). The Geogebra resource Completing the square uses area to show what is happening when you complete the square. There are also videos on Solving quadratics and Completing the square.

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MEI C1 Algebra Section 2 Study plan 7. Exercise 1D Try the first, a middle and the last part of questions 6 to 8, then do about half of the longer questions. You may download worked solutions to questions 8*, 9*, 11*, 13*, 15*. 8. For additional practice, use the interactive questions Solving quadratics by factorisation and Forming and solving quadratics. 9. You can also try the Quadratics puzzle.

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MEI Core 1 Basic Algebra Section 2: Quadratics Notes and Examples These notes contain subsections on  Factorising quadratic expressions  Graphs of quadratic functions  Solving quadratic equations by factorisation  Solving quadratic equations using the formula  Problem solving

In this section of work you will be studying quadratic functions and their graphs. You probably already know something about this topic, but you will now be taking it a little further.

Factorising quadratic expressions Again, this should be revision of GCSE work. It is essential that you are confident in factorisation.

Example 1 Factorise the expressions (i) x² + 4x + 3 (ii) x² – 4x – 12 (iii) 2x² – 7x + 6 Solution (i) x² + 4x + 3 = (x …..)(x …...)

(ii)

Start with an x in each bracket

x² + 4x + 3 = (x + 1)(x + 3)

You need two numbers whose sum is 4 and whose product is 3. These are +1 and +3.

x² – 4x – 12 = (x …..)(x …..)

Start with an x in each bracket

x² – 4x – 12 = (x – 6)(x + 2)

You need two numbers whose sum is –4 and whose product is –12. These are –6 and +2.

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MEI C1 Algebra Section 2 Notes and Examples (iii) 2x² – 7x + 6 = (2x …..)(x …..) 2x² – 7x + 6 = (2x – 3)(x – 2)

In this case you need to start with 2x in one bracket and x in the other.

It is not so straightforward to find the two numbers in this case, because of the 2x in one bracket. The two numbers must have a product of +6, and as the coefficient of x is negative, they must both be negative. Try the different possibilities (–1 and –6, or –2 and –3, in either order), until you find the correct one.

You can see step-by-step examples of factorising quadratics in this PowerPoint presentation. You can also look at the Factorising quadratics video. For practice in examples like the ones above, try the interactive questions Factorising quadratics. You can also test yourself using the Flash resource Factorising quadratics.

Sometimes algebraic expressions which look quite complicated can be simplified by factorising.

Example 2

It can be tempting to try to “cancel” x² from the top and the bottom. Don’t! You can only cancel something which is a factor of the top and the bottom.

x2 1 Simplify 2 . x  2x  3 Solution x2 1 ( x  1)( x  1)  2 x  2 x  3 ( x  3)( x  1)

x 1  x3

You can now cancel the factor (x – 1) from the top and the bottom.

Graphs of quadratic functions Factorising a quadratic expression gives you information about the graph of a quadratic function. Do not think of this work as just algebraic manipulation, think about it also in terms of the graph of the function. Linking algebra and graphs is a very important mathematical skill; the good news is that being able to consider problems both algebraically and graphically usually makes them easier! A graphic calculator or computer package will be very useful.

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MEI C1 Algebra Section 2 Notes and Examples You may already be familiar with the graph of the simplest quadratic function, y = x².

y = x²

The curve given by graphs of quadratics is called a parabola. Notice that all quadratic graphs have reflection symmetry. The mirror line is always a vertical line through the turning point or vertex of the curve (shown in yellow on these graphs).

All other quadratic graphs have basically the same shape, but they may be “stretched”, “squashed”, shifted or inverted.

y = x² – x – 2 y = 3 – 2x – x²

Factorise the equations of these graphs. What is the relationship between the factorised form and the graph? Can you explain this?

Notice that the graphs of functions with a negative x² term are inverted (upside down). You will look at the graphs of quadratic expressions in more detail in Polynomials Section 3.

Solving quadratic equations by factorisation Solving quadratic equations is important not just from the algebraic point of view, but because it gives you information about the graph of a quadratic function. The solutions of the equation ax² + bx + c = 0 tells you where the graph of the function y = ax² + bx + c crosses the x-axis, since these are the points where y = 0.

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MEI C1 Algebra Section 2 Notes and Examples Some quadratic equations can be solved by factorising.

Example 3 Solve these quadratic equations by factorising. (a) x² + 2x – 8 = 0 (b) 2x² + 11x + 12 = 0 Solution (a) x² + 2x – 8 = 0 (x + 4)(x – 2) = 0 x + 4 = 0 or x – 2 = 0 x = –4 or 2 (b)

For this expression to be zero, one or other of the factors must be zero.

2x² + 11x + 12 = 0 (2x + 3)(x + 4) = 0 2x + 3 = 0 or x + 4 = 0 x =  32 or –4

You can see further examples using the Flash resources Quadratic equations 1 (in which the coefficient of x² is always 1) and Quadratic equations 2 (in which the coefficient of x² is greater than 1). For practice in examples like the ones above, try the interactive questions Solving quadratics by factorisation.

Solving quadratic equations using the formula It is possible to solve quadratic equations using the method of completing the square (see pages 19 – 20 of the textbook); however the quadratic formula is just a generalisation of this method. Pages 20 – 21 of the textbook show how the quadratic formula is derived from the technique of completing the square. To help you understand this technique, you can look at the Flash resource Quadratic equations 4, and the Completing the square video. You may also find the Geogebra resource Completing the square useful – this uses area to show what is happening when you complete the square. However, you should understand that completing the square is not normally used to solve a quadratic equation, since the quadratic formula is quicker and easier to use. Completing the square is covered in greater detail in Polynomials Section 3, where it is used to learn more about the graphs of quadratic functions . The quadratic formula for the solutions of the equation ax²  bx  c  0 is b  b²  4ac x . 2a

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MEI C1 Algebra Section 2 Notes and Examples The expression b²  4ac is called the discriminant. This is very important as it tells you something about the nature of the solutions. In each case the solution(s) correspond to the points where the graph meets the x-axis. 

If the discriminant is positive, then there are two real solutions. (If the discriminant is a positive square number, then the two real solutions are rational and it is possible to solve the equation by factorisation; otherwise the solutions are irrational and you must use the quadratic formula.)

y = x² + 2x – 3 Discriminant = 16 Two rational solutions 

y = x² + x – 3 Discriminant = 13 Two real, irrational solutions

If the discriminant is zero, then the quadratic is a perfect square and there is one real solution, which can be found by factorisation. y = x² + 2x + 1 Discriminant = 0 One real solution



If the discriminant is negative, then there are no real solutions. y = x² + 2x + 2 Discriminant = -4 No real solutions As the graph does not meet the x-axis, there cannot be any real solutions.

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MEI C1 Algebra Section 2 Notes and Examples When you need to solve a quadratic equation, it is useful to quickly work out the discriminant before you start, so that you know whether there are real solutions, and whether the equation can be solved by factorisation. Note that when you use the quadratic formula in C1, you will need to give answers in an exact form, involving a square root, since C1 is a noncalculator paper. Sometimes you may need to simplify square roots, e.g. 8  4  2  2 2 , so that you can cancel down fractions. You will learn more about working with square roots in chapter 5.

You can look at the relationship between the discriminant of a quadratic equation and the number of roots using the Flash resource The discriminant. You can also see examples of solving quadratic equations using the formula in the Flash resource Quadratic equations 3. The Solving quadratics video looks at solving quadratics by all the methods covered. Try the Quadratics puzzle, either on your own or with one or two others. Cut out all the pieces and match up each equation with its solution. The pieces will form a large hexagon.

Example 4 For each of the following quadratic equations, find the discriminant and solve the equation, where possible, by a suitable method 6 x²  11x 10  0 (i) 2 x²  5x  1  0 (ii) 4 x²  12 x  9  0 (iii) 3x²  2 x  4  0 (iv) Solution (i) a = 2, b = -5, c = 1 Discriminant = (5)²  4  2 1  25  8  17 Since the discriminant is positive, there are two real solutions. As it is not a square number, the equation must be solved using the quadratic formula. b  b ²  4ac x 2a

5  17 2 2 5  17  4 

(ii) a = 6, b = 11, c = -10 Discriminant = 11²  4  6  10  121  240  361 Since the discriminant is positive, there are two real solutions. As it is a square

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MEI C1 Algebra Section 2 Notes and Examples number (19²), the equation can be solved by factorisation. 6 x ²  11x  10  0

(3x  2)(2 x  5)  0 x  23 or x   52 (iii) a = 3, b = -2, c = 4 Discriminant = (2)²  4  3  4  4  48  44 Since the discriminant is negative, there are no real solutions. (iv) a = 4, b = 12, c = 9 Discriminant = 12²  4  4  9  144 144  0 Since the discriminant is zero, there is one solution and the equation can be solved by factorisation into a perfect square. 4 x ²  12 x  9  0

(2 x  3)²  0 x   32

Problem solving Some problems, when translated into algebra, involve quadratic equations.

Example 5 A rectangular box has width 2 cm greater than its length, and height 3 cm less than its length. The total surface area of the box is 548 cm². What are the dimensions of the box? Solution Let the length of the box be x cm. The width of the box is x + 2 cm, and the height is x – 3 cm. The surface are of the box is given by 2x(x + 2) + 2x(x – 3) + 2(x + 2)(x – 3) Divide through by 2 2x(x + 2) + 2x(x – 3) + 2(x + 2)(x – 3) = 548 x(x + 2) + x(x – 3) + (x + 2)(x – 3) = 274 x² + 2x + x² – 3x + x² – x – 6 = 274 The discriminant is 3364, which is 58², 3x² – 2x – 280 = 0 so this must factorise (3x + 28)(x – 10) = 0 x = 10

3x + 28 = 0 gives a negative value of x, which does not make sense in this context. So the solution must be x – 10 = 0. The length of the box is 10 cm, the width is 12 cm and the height is 7 cm.

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MEI C1 Algebra Section 2 Notes and Examples Notice that in Example 5, you could discard one of the possible solutions as a negative solution did not make sense in the context. This is not always the case. In some situations, a negative solution can have a practical meaning. For example if the height of a stone thrown from the edge of a cliff is negative, this simply means that the stone is below the level of the cliff at that point. However, if the stone was thrown from level ground, then a negative height does not make sense. Some problems leading to quadratic equations do have two possible solutions. Always consider whether your solution(s) make sense in the context. For practice in examples like the ones above, try the interactive questions Forming and solving quadratics.

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Core 1 Basic Algebra Section 2: Quadratics Crucial points 1. Make sure that you can multiply out and factorise confidently These algebraic skills are vital in many areas of mathematics at this level. Practise them until you can do them confidently. 2. Remember the relationship between the solutions of a quadratic equation and the corresponding quadratic graph The solutions of a quadratic equation tell you where the corresponding quadratic graph crosses the x-axis. This is very useful in sketching quadratic graphs. 3. Remember how to calculate the discriminant and what information it gives you Remember that the discriminant of a quadratic equation gives you useful information about the nature of its solutions, so it is often useful to work out the discriminant before you try to solve the equation. 4. When solving problems, make sure your answer makes sense Always look at your answers to problems in the light of the original question. If there are two solutions, do they both make sense, or should one be discarded? Think carefully about the meaning of a negative solution, since this may or may not be a valid solution.

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MEI Core 1 Basic Algebra Section 2: Quadratics Exercise Do not use a calculator for this exercise. 1. Factorise these quadratic expressions. (i) x² + 5x + 6 (ii) x² + x – 12 (iv) x² – 6x + 8 (v) 2x² + 3x + 1 (vii) 4x² – 8x + 3 (viii) 4x² – 25

(iii) (vi) (ix)

x2 − 9 3x² + x – 2 6x² – x – 12

2. Simplify these expressions where possible. x2 + x − 6 x2 − 4 x + 4 (ii) (i) x2 − x − 2 x2 + x − 6 x2 + x − 2 4x2 −1 (iii) (iv) x2 + 4 x + 3 4x2 − 4x − 3 2x + 3 x+2 x2 − x − 6 × ( 3x 2 − 2 x − 1) ÷ (v) (vi) 3x + 1 2 x2 − x − 1 2x +1

3. Solve these quadratic equations by factorising. (i) x² + 4x + 3 = 0 (ii) x² + 5x – 6 = 0 (iii) x² – 6x + 8 = 0 (iv) x² – 7x – 18 = 0 (v) 2x² + 5x + 3 = 0 (vi) 2x² + x – 6 = 0 (vii) 4x² – 3x – 10 = 0 (viii) 6x² – 19x + 10 = 0 4. Solve the following quadratic equations, where possible. Give answers in exact form. (i) x² + 2x – 2 = 0 (ii) x² – 3x + 5 = 0 (iii) 2x² + x – 4 = 0 (iv) 2x² – 5x – 12 = 0 (v) x² – 5x – 3 = 0 (vi) 3x² + x + 1 = 0 (vii) 4x² + 12x + 9 = 0 (viii) 4x² + 10x + 5 = 0 5. The length of a rectangle is 3 cm greater than its width. The area of the rectangle is 40 cm². Find the length and width of the rectangle.

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MEI Core 1 Basic Algebra Section 2 : Quadratics Solutions to Exercise 1. (i)

x 2  5 x  6  x 2  3x  2 x  6  x( x  3)  2( x  3)  ( x  2)( x  3)

(ii) x 2  x  12  x 2  4 x  3 x  12

 x( x  4)  3( x  4)  ( x  3)( x  4) (iii) x 2  9  ( x  3)( x  3) (iv) x 2  6 x  8  x 2  2 x  4 x  8

 x( x  2)  4( x  2)  ( x  4)( x  2) (v) 2 x 2  3 x  1  2 x 2  x  2 x  1

 x(2 x  1)  1(2 x  1)  ( x  1)(2 x  1) (vi) 3 x 2  x  2  3 x 2  3 x  2 x  2

 3 x( x  1)  2( x  1)  (3 x  2)( x  1) (vii) 4 x 2  8 x  3  4 x 2  2 x  6 x  3

 2 x(2 x  1)  3(2 x  1)  (2 x  3)(2 x  1) (viii)

4 x 2  25  (2 x  5 )(2 x  5 )

(ix) 6 x 2  x  12  6 x 2  8 x  9 x  12

 2 x(3 x  4)  3(3 x  4)  (2 x  3)(3 x  4) 2. (i)

x 2  x  6 ( x  3)( x  2) x  3   x 2  x  2 ( x  2)( x  1) x  1

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MEI C1 Algebra Section 2 Exercise solutions x 2  4x  4 ( x  2) 2 x 2 (ii)   2 x  x  6 ( x  3)( x  2) x  3 (iii)

x 2  x  2 ( x  2)( x  1)  x 2  4 x  3 ( x  3)( x  1)

This expression cannot be simplified. (iv)

(2 x  1)(2 x  1) 2 x  1 4x 2  1   2 4 x  4 x  3 (2 x  1)(2 x  3) 2 x  3

(v)

2x  3 2x  3  3x 2  2 x  1   (3 x  1)( x  1)  (2 x  3)( x  1) 3x  1 3x  1

(vi)

x 2 x2  x 6 x 2 ( x  3)( x  2)    2 2x  x  1 2x  1 (2 x  1)( x  1) 2x  1 x 2 2x  1   (2 x  1)( x  1) ( x  3)( x  2) 

3. (i)

1 ( x  1)( x  3)

x 2  4x  3  0 ( x  3)( x  1)  0 x  3 or x  1

(ii) x 2  5 x  6  0

( x  6)( x  1)  0 x  6 or x  1 (iii) x 2  6 x  8  0

( x  2)( x  4)  0 x  2 or x  4 (iv) x 2  7 x  18  0

( x  9)( x  2)  0 x  9 or x  2 (v) 2 x 2  5 x  3  0

(2 x  3)( x  1)  0

x   23 or x  1

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MEI C1 Algebra Section 2 Exercise solutions (vi) 2 x 2  x  6  0

(2 x  3)( x  2)  0

x

or x  2

3 2

(vii) 4 x 2  3 x  10  0

(4 x  5 )( x  2)  0

x   54 or x  2 (viii) 6 x 2  19 x  10  0

(3 x  2)(2 x  5 )  0

x

2 3

or x 

5 2

4. (i) x 2  2 x  2  0

a  1, b  2, c  2 Discriminant  b 2  4ac  2 2  4  1  2  4  8  12 x

b  b 2  4ac 2  12 2  2 3    1  3 2a 2 2

(ii) x 2  3 x  5  0

a  1, b  3, c  5 Discriminant  b 2  4ac  ( 3)2  4  1  5  9  20  11 The discriminant is negative so there are no real roots. (iii) 2 x 2  x  4  0

a  2, b  1, c  4 Discriminant  b 2  4ac  1 2  4  2  4  1  32  33

x

b  b 2  4ac 1  33 1  33   2a 22 4

(iv) 2 x 2  5 x  12  0

a  2, b  5 , c  12 Discriminant  b 2  4ac  ( 5 )2  4  2  12  25  96  121 Since the discriminant is a perfect square, then it is possible to factorise. (2 x  3)( x  4)  0

x   23 or x  4 (v)

x2  5 x 3  0 a  1, b  5 , c  3 Discriminant  b 2  4ac  ( 5 )2  4  1  3  25  12  37

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MEI C1 Algebra Section 2 Exercise solutions x

b  b 2  4ac 5  37  2a 2

(vi) 3 x 2  x  1  0

a  3, b  1, c  1 Discriminant  b 2  4ac  1 2  4  3  1  1  12  11 The discriminant is negative so there are no real roots. (vii) 4 x 2  12 x  9  0

a  4, b  12, c  9 Discriminant  b 2  4ac  12 2  4  4  9  144  144  0 Since the discriminant is zero, there is a repeated root and the equation can be factorised. (2 x  3)2  0

x   23 (viii) 4 x 2  10 x  5  0

a  4, b  10, c  5 Discriminant  b 2  4ac  102  4  4  5  100  80  20

x

b  b 2  4ac 10  20 10  2 5 5  5    2a 24 8 4

5. Let x be the width of the rectangle, so the length is x + 3. Area  x( x  3) x( x  3)  40

x 2  3 x  40 x 2  3 x  40  0 ( x  8)( x  5 )  0 x  8 or 5

Dimensions must be positive, so width is 5 cm and length is 8 cm.

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Quadratic equations Hexagonal Jigsaw The three diagrams on the following pages will ultimately fit together to form a large hexagon. Before you start, the three diagrams must be cut along the lines to make twenty-four equilateral triangles. For the triangles to be fitted together, you must match a quadratic equation to its solution. To build up the puzzle, place together the edges on which a quadratic and its solution are written, so that the triangles are joined along this edge. Begin by just finding pairs of matching quadratics and solutions, and placing them edge-to-edge. As you progress you will find that all of the pieces will eventually link up to form a large hexagon. By all means try this by yourself, but it is really designed to be a group activity. Working with other students will help to highlight any misconceptions you may have, as well as making the task more enjoyable. At first glance this may appear to be an easy undertaking, but you will find it takes quite a lot of thought and errors can be made very easily. Make sure you check each coupling, or you could be left with one or two pieces that appear to fit nowhere.

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x

−3

x +2

+

=2 x

0 = 1

3 x 2 + 3x − 2 = x 2

±1

x

x=

x = 0, x = 3 2

2

2

5x

x=

4 −2

x=

−5

x = x

−1

x 2 − 2x + 1 = 0

2

x

1 2

=0

x

+

x x 2 − 4x = − 3

2

3 = 2

x= 0,

x = 2, x =

−6

x2 −9 = 0

x

±

2

−1

x

± 2 =

=

−

6

x

−x

x=2

2

4=

x = ±2

4x +

x

, −3

2

2=

x=

1 2 x = − 2, x =

, 1 =

0

−1

x= ± 3

x=1

x 2

x

=0

2

2

+9

x = 1±

x

x=

=0

± 2 −

6

=0

3 = x

−2 x−

x 2 + 2x = 0

7−7x2 = 0

x−

0 = 2

2

x=

+4

x

=

− 1 2 ,

2

−

x

16

x

4x + x 2 + 3 = 0

x

+3

x= 1 3

x = 1, x = 2 2

x+

x= 3

x 2 − 2 = 4x

x= −2 ,

1=

0

x=

6x

2

x = − 3, x = 1 2

e non

− 1 ,

+x

x = 3

−1 =0

2x2 −8 = 0

x − 3 x = 0

x 2 + 2x − 1 = 0

=0

−6

+x

5 x 2 + 25x = 0 x 2

4x

+4

2

=4

− 4 x+ 1 = 0

x = − 1, x = − 2

x=

2x

2

x

−2

2

x +4

−

0 = 6

x= 1 2

x = 0, x = − 2

x=

x = − 3, x = 2

2x

− 2

x= 0

x= 5 x= 0,

±4

+ x 5

0 = 2

x

MEI Core 1 Basic Algebra Section 2: Quadratics Multiple Choice Test Do not use a calculator for this test. 1) The expression (2x – 5)(x + 3) is equivalent to (a) 2x² + x – 15 (c) 2x² + 11x – 25 (e) I don’t know

(b) 2x² – x – 15 (d) 2x² – 2x – 15

2) The expression 4y² + 5y – 6 is equivalent to (a) (4y – 1)(y + 6) (c) (2y – 1)(2y + 6) (e) I don’t know

3) The expression

(b) (2y – 3)(2y – 2) (d) (4y – 3)(y + 2)

y2 1 can be simplified to y2  y  2

y 1 y2 1 (c) y2 (e) I don’t know (a)

(b)

y 1 y2

(d) cannot be simplified

4) The discriminant of the quadratic equation 2x² + 5x – 1 = 0 is (a) 17 (c) 27 (e) I don’t know

(b) 33 (d) –3

5) The quadratic equation in Question 4 has (a) two real irrational solutions (c) one real solution (e) I don’t know

(b) two rational solutions (d) no real solutions

© MEI, 21/12/09

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MEI C1 Algebra Section 1 MC test 6) The solutions of the equation x² + 2x – 5 = 0 are (a) no real solutions (c) 1  12 (e) I don’t know

(b) 1  6 (d) 1  24

7) The solutions of the equation 2x² – 11x + 15 = 0 are (a) no real solutions 11  241 (c) 4 (e) I don’t know

(b) 1.5 and 5 (d) 2.5 and 3

8) The solutions of the equation 3x² – 2x + 4 = 0 are (a) no real solutions

(b) 2 and

1  13 3 (e) I don’t know

(d)

(c)

2 3

1  11 3

9) The solutions of the equation 2x² + 5x – 4 = 0 are

5  57 4 5  57 (c) 4 (e) I don’t know (a)

5 7 4 5  7 (d) 4 (b)

10) The quadratic equation x2  kx  2k  3  0 has equal roots. The possible value(s) of k are (a) 0 only (c) 3 or 4 (e) I don’t know

(b) 2 only (d) 2 or 6

© MEI, 21/12/09

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