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Quadratic equations Test A
Name: ___________________________
Section A Multiple Choice
1
Which of the following is not a
quadratic equation?
x 2  4x  6  0
A
x( x  2)  4 x  1
B
C
D
2
3
4
5
C
6
The solutions to ( x  3) 2  16 are:
A x = 4, x = –4
B x = 7, x = –1
C x = 1, x = –7
D none of the above.
B
7
The solutions to the equation
x 2  6 x  5  0 are:
A x = 1 and x = 5
B x = 1 and x = 5
C x = 1 and x = 5
D x = 1 and x = 5
A
8
The equation x 2  6 x  7  0 has:
A two rational solutions
B two irrational solutions
D one irrational solution
E no solutions
B
9
Which of the following quadratic
equations has exactly one solution?
A x 2  8 x  16  0
B 2x2  7x  0
C x 2  8x  9  0
D x2  4  0
A
10
The quadratic equation x 2  1  0 has
no roots because:
A the square root of 1 is +1 or –1
B it is not a quadratic trinomial
C you cannot take the square root of
–1
D 1 is not a prime number.
C
( x  2)( x 2  2)  0
( x  2)( x  2)  0
The solutions to (3x – 4)(5x – 8) = 0
are:
A x = 4, x = 8
B x = –4, x = –8
1
5
C x=1 ,x=
8
3
1
3
D x =1 , x = 1
3
5
D
The solutions to 4 x 2  36  0 are:
A x = 6, x = –6
B x=6
C x=3
D x = 3, x = –3
D
The solutions to 5 x 2  20 x  0 are:
A x = 2 or x = 0
B x = 2 or x = 0
C x = 4 or x = 0
D x = 4 or x = 0
D
The solutions to x 2  13 x  42  0 are:
A 7 and 7
B 7 and 6
C 7 and –6
D 7 and 6
C
Maths Quest 10 for New South Wales 5.3 pathway
Chapter 4
1
Solving quadratic equations Test A
Name: ___________________________
Section B Short/Extended answer
1
2
Identify the following equations as linear,
quadratic or other.
(a) 5 x  5 x 4  3
(b)
x  7 x 2  8x 2  3
(c)
2x  3  0
(d)
8y 2  3y  7  0
(a)
The highest power in the equation is 4, so
this is neither quadratic nor linear.
(b)
The highest power in the equation is 2, so
this is quadratic.
(c)
The highest power in the equation is 1, so
this is linear.
(d)
The highest power in the equation is 2, so
this is quadratic.
Solve the following quadratic equations that are
expressed in factorised form.
(a) (x – 6)(x + 2) = 0
(a)
(b) (2x +
3
4
1
)( –5x + 40) = 0
3
1
)( –5x + 40) = 0
3
1
(2x + ) = 0
or (–5x + 40) = 0
3
1
2x = –
or
–5x = –40
3
1
x =–
or
x =8
6
6
(a)
8 x 2  32  0
Maths Quest 10 for New South Wales 5.3 pathway
(x – 6)(x + 2) = 0
x–6=0
or x + 2 = 0
x=6
or
x = 2
(b) (2x +
Solve the following quadratic equations.
(a) x 2  81  0
(b)
3
(b)
Chapter 4
x – 81
(x + 9)(x – 9)
x+9
x
2
=0
=0
= 0 or x – 9 = 0
= –9 or
x =9
8x 2 – 32
8( x 2  4)
8(x + 2)(x – 2)
x+2
x
=0
=0
=0
=0
= –2
or x – 2 = 0
or x = 2
2
4
Solve the following quadratic equations.
(a) x 2  9 x  0
4
(a)
x  9x  0
2
x( x  9)  0
x  0 or x  9  0
x  0 or
(b)
1 2 4
x  x0
6
7
(b)
x9
1 2 4
x  x0
6
7
1
4
x( x  )  0
6
7
1
4
x  0
6
7
1
4
or
x
6
7
x  0 or
x0
5
Solve the following quadratic equations.
(a) x 2  3x  10  0
x0
or
x  6
x0
or
x
4
7
24
7
6
(a)
x  3x  10  0
2
( x  5)( x  2)  0
x  5  0 or x  2  0
x  5 or
(b)
3x 2  19 x  14  0
(b)
x2
3 x 2  19 x  14  0
(3 x  2)( x  7)  0
3 x  2  0 or x  7  0
3x  2
or
x  7
2
3
or
x  7
x
Maths Quest 10 for New South Wales 5.3 pathway
Chapter 4
3
6
Solve each of the following quadratic
equations:
(a) 2 x 2  13x  15  0
6
2 x 2  13 x  15  0
(a )
2 x 2  10 x  3 x  15  0
2 x( x  5)  3( x  5)  0
( x  5)( 2 x  3)  0
x5  0
or 2 x  3  0
(b)
6 x 2  5x  6  0
x5
or
2x  3
x5
or
x 1
1
2
6 x 2  5x  6  0
( b)
6x 2  9x  4x  6  0
3 x ( 2 x  3)  2( 2 x  3)  0
( 2 x  3)(3 x  2)  0
2x  3  0
or 3 x  2  0
2 x  3
x  1
7
1
2
or
3x  2
or
x
Solve the following by using the ‘complete the
square’ method. Give an exact answer.
2
3
3
The coefficient of x is 6.
x  6x 1  0
2
2
1 
2
  6  3  9
2 
2
x  6x  9  1 9
( x  3) 2  10
x  3   10
x  3  10 or x  3  10
Maths Quest 10 for New South Wales 5.3 pathway
Chapter 4
4
8
Let x be an unknown positive number.
Translate the various phrases into algebraic
When double its value is squared and 5 times
terms.
its value removed, then it is equal to 3 times the
(2 x) 2  5 x  3x 2  6
square of the original number plus 6. Find x.
4 x 2  5 x  3x 2  6
6
x 2  5x  6  0
( x  6)( x  1)  0
x  6  0 or
x  6 or
x 1  0
x  1
The solution is x = 6. The solution x = 1 is
rejected as we are told in the question that x is
positive.
9
Solve the following quadratic equations using
the quadratic formula. Where necessary give
your answer correct to 2 decimal places.
(a) x 2  9 x  14  0
6
(a ) x 2  9 x  14  0
x
 b  b 2  4ac
2a
9  (9) 2  4  1  14

2
9  25

2
9  25
9  25
x
or
2
2
x7
x2
(b)
(b) x 2  8 x  1  0
x 2  8x  1  0
x
 b  b 2  4ac
2a
 8  8 2  4  1  ( 1)
2
 8  68

2
 8  68  8  68
x
or
2
2
x  0.12 or x  8.12

Maths Quest 10 for New South Wales 5.3 pathway
Chapter 4
5
10
a = 1, b = 9, c = 15
For the equation x 2  9 x  15  0 use the
discriminant to determine whether the equation   b 2  4ac
has:
 9 2  4  1  (15)
 two rational solutions
 81  60
 two irrational solutions
 one solution
 141
 no real solutions.
As  > 0 the equation has two solutions.
As  is not a perfect square the solutions are
irrational.
3
x 2  9 x  15  0 has two irrational solutions.
11
Solve the equation 2( x  5) 2  ( x  5)  15  0.
2( x  5) 2  ( x  5)  15  0
Let u = x – 5
2u 2  u  15  0
3
(2u  5)(u  3)  0
5
u =  or u = 3
2
5
 x – 5 =  or x – 5 = 3
2
5
that is x =
or
x =8
2
Maths Quest 10 for New South Wales 5.3 pathway
Chapter 4
6