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Bks MaaSL 0306 ws00 Xxaann

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97 views3 pages

Bks MaaSL 0306 ws00 Xxaann

Uploaded by

Aanya Ralhan
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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Additional exercise

3.6 The quadratic formula and the


discriminant
1 Use the quadratic formula to find the roots of each equation.

a 4x2  x  2  0 b 2x2  3x  1  0 c x2  4x  2  0

d 5x2  x  2  0 e 3x  x2  5 f 2x2  3x  2

g x  x  2  6 x  5 h x 3x  4  1 i 5x2  9x  10  2x  4

2 Identify which of the equation(s) in question 1 could be solved by factorization and then show
the factorization.

5  57
3 The zeros of the function f  x   2x2  5x  c , where c is a real number, are
4
5  57
and . Find c.
4

4 Find the value of the discriminant and then state the nature of the roots of each equation.

a 2x2  x  4  0 b 3x2  4x  4  0 c 4x2  12x  9  0

d 3x  5  x2 e 2x2  8x  5  0 f 5x2  2x  3

5 Identify which of the equation(s) in question 4 have rational roots. Justify your answer.

6 Find the value(s) of k such that the equation 2x2  4x  k  0 has two distinct real roots.

7 Find the value(s)of p such that the graph of f  x   x2  2px  3p lies on the x -axis.

8 Find the value(s)of p such that the vertex of the graph of f  x   2x2  4x  m  1 has
no x -intercepts.

9 Solve each inequality.

a x2  25  0 b 2x2  x  6  0 c x2  2x  5

10 Find the value(s) of k that make the statement true.

a x2  3kx  1  0 has two distinct real roots

b the graph of f  x   kx2  2kx  3  2k has no x -intercepts

© Oxford University Press 2019 Additional exercise 1


Additional exercise

Answers

  1   1  4  4  2


2
2 1  33
1 a 4x  x  2  0  x  
2  4 8

3  32  4 2 1 3  1 1
b 2 x 2  3x  1  0  x    1, 
2 2  4 2

  4   4  4 1 2


2
4 8
c x 2  4x  2  0  x   or 2  2
2 1 2

  1   1  4 5 2


2
1  39
d 5x 2  x  2  0  x    no real roots
2 5 10

3  32  4  1 5 3  29 3  29
e 3x  x 2  5   x 2  3x  5  0  x   or
2  1 2 2

3  32  4 2  2 3  5 1
f 2x 2  3x  2  2x 2  3x  2  0  x    2,
2 2  4 2

  8   8  4 1 5


2
8  44
g x  x  2  6 x  5  x 2  8x  5  0  x   or 4  11
2 1 2

4  42  4 3  1 4  28 2  7
h x  3x  4   1  3x 2  4 x  1  0  x   or
2  3 6 3

  11   11  4 5 6 


2
11  1 6
i 5x 2  9x  10  2x  4  5x 2  11x  6  0  x    1,
2 5 10 5

1
2 b 2x2  3x  1  0  2x  1  x  1  0  x   , 1
2

1
f 2x2  3x  2  2x 2  3x  2  0  2x  1  x  2  0  x  , 2
2
6
i 5x2  9x  10  2x  4  5x2  11x  6  0  5x  6   x  1  0  x  ,1
5

  5   5  4 2  c 


2
5  25  8c
3 2 x 2  5x  c  0  
2 2  4
5  25  8c 5  57
  25  8c  57  8c  32  c  4
4 4

4 a   12  4 2 4  31 ; no real roots

b   42  4 3 4  64 ; two distinct real roots

c   122  4  49  0 ; two equal real roots (one repeated root)

d 3x  5  x2   x2  3x  5  0;   32  4  15  29 ; two distinct real roots

   8  4 2 5  24 ; two distinct real roots


2
e

   2  4 5 3  56 ; no real roots


2
f

5 b 3x2  4x  4  0 has rational roots because 64 is a perfect square.

c 4x2  19x  9  0 has rational roots because 0 is a perfect square.

© Oxford University Press 2019 2


Additional exercise

6 42  4 2 k   0  16  8k  0  8k  16  k  2

7 vertex on x -axis  x2  2px  3p  0 has one repeated real root, so   0 .

 2p  4 13p  0  4p2  12p  0  4p  p  3  0  p  0,3


2

8 no x -intercepts  2x2  4x  m  1  2x2  4x   m  1  0 has no real roots, so   0 .

42  4 2 m  1  0  16  8m  8  0  8m  24  m  3

9 a x2  25  0   x  5 x  5  0  x  5,5

so x2  25  0  5  x  5

3
b 2 x 2  x  6  0   2 x  3  x  2   0  x   ,2
2

3
so 2x2  x  6  0  x   or x  2
2

2  22  4  1 5
c  x 2  2x  5  0  x  1 6
2  1

so x2  2x  5  1  6  x  1  6

2 2
 3k   4 1 1  0  9k 2  4  0  3k  2 3k  2  0  k  
2
10 a or k 
3 3

b no x -intercepts  kx2  2kx  3  2k   0 has no real roots

 2k   4  k  3  2k   0  4k 2  12k  8k 2  0  4k  k  3  0  k  3 or k  0


2

© Oxford University Press 2019 3

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