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SPM Maths

Quadratic Equations Long Questions (Question 14 & 15)

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Question 14 (4 marks):
An aquarium has the length of (x + 7) cm, the width of x cm and the height of 60 cm.
The total volume of the aquarium is 48000 cm3. The aquarium will be filled fully with water.
Calculate the value of x.

Solution:


Volume of aquarium=48000  cm 3 So, ( x )( x+7 )( 60 )  cm 3 =48000  cm 3 ( x )( x+7 )= 48000 60 x 2 +7x=800 x 2 +7x−800=0 ( x−25 )( x+32 )=0 x−25=0   or   x+32=0 x=25   or   x=−32( not accepted ) So the value of x=25 cm



Question 15 (4 marks):
Diagram 3 shows a garden path with a rectangular shape. There are 8 similar circular stepping stone built in the path.


Given the area of the path is 32 m2, find the diameter, in m, of one piece of stepping stone.

Solution:
Given the area of the path =32  m 2 =320000  cm 2 x( x+4 )=320000 x 2 +4x−320000=0 a=1, b=4, c=−320000 x= −b± b 2 −4ac 2a x= −4± 4 2 −4( 1 )( −320000 ) 2( 1 ) x= −4± 1280016 2 x= −4+1131.38 2    or    −4−1131.38 2 x=563.69   or   −567.69 ( ignored ) Diameter of one piece of stepping stone=x =563.69 cm =5.64 m

2.5 SPM Practice (Long Questions)

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Question 13 (12 marks):
(a) Complete Table 2 in the answer space, for the equation y= 30 x  by writing down the values of y when x = 2 and x = 5.

(b) For this part of the question, use the graph paper. You may use a flexible curve ruler.
Using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the y-axis, draw the graph of  y= 30 x  for 0≤x≤7.  

(c) From the graph in 12(b), find
(i) the value of y when x = 2.6,
(ii) the value of x when y = 17.5.

(d) Draw a suitable straight line on the graph in 12(b) to find the values of x which satisfy the equation  30 x =−5x+30 for 0≤x≤7.
State the values of x.

Answer:
Table 2


Solution:
(a)



(b)



(c) From the graph
(i) When x = 2.6; y = 11.5
(ii) When y = 17.5; x = 1.7

(d)
Given, 30 x =−5x+30   y=−5x+30 From the graph; x=1.25 and 4.75



2.5 SPM Practice (Long Questions)

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Question 12 (12 marks):
(a) Complete Table 2 in the answer space, for the equation y = –x3 + 4x + 10 by writing down the values of y when x = –2 and x = 1.5.

(b) For this part of the question, use the graph paper. You may use a flexible curve ruler.
Using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 10 units on the y-axis, draw the graph of y = –x3 + 4x + 10 for –3 ≤ x ≤ 4.

(c) From the graph in 12(b), find
(i) the value of y when x = –2.5,
(ii) the positive value of x when y = 4.

(d) Draw a suitable straight line on the graph in 12(b) to find the values of x which satisfy the equation x3 – 14x + 5 = 0 for –3 ≤ x ≤ 4.
State the values of x.

Answer:




Solution:
(a)
y = –x3 + 4x + 10
When x = –2
y = –(–2)3 + 4(–2) + 10
y = 8 – 8 + 10
y = 10

When x = 1.5
y = –(1.5)3 + 4(1.5) + 10
y = –3.375 + 6 + 10
y = 12.625

(b)




(c) From graph
(i) When x = –2.5; y = 16
(ii) When y = 4; x = 2.5

(d)
x3 – 14x + 5 = 0
–x3 + 14x – 5 = 0
–x3 + (4x + 10x) + (10 – 15) = 0
–x3 + 4x + 10 = –10x + 15
Thus, y = –10x + 15


From graph, the values of x are 0.35 and 3.5.


2.5 SPM Practice (Long Questions)

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Question 11 (12 marks):
(a) Complete Table 12 in the answer space, for the equation y = –x2 + 2x + 10 by writing down the values of y when x = –1 and x = 2.

(b) For this part of the question, use the graph paper. You may use a flexible curve ruler.
Using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 2 units on the y-axis, draw the graph of y = –x2 + 2x + 10 for –3.5 ≤ x ≤ 4.

(c) From the graph in 12(b), find
(i) the value of y when x = –1.5,
(ii) the positive value of x when y = 8.2.
[adinserter block="3"]

(d) Draw a suitable straight line on the graph in 12(b) to find the values of x which satisfy the equation 7 – x = x2 for –3.5 ≤ x ≤ 4.
State the values of x.

Answer:



Solution:
(a)
y = –x2 + 2x + 10
When x = –1
y = –(–1)2 + 2(–1) + 10
y = –1 – 2 + 10
y = 7

when x = 2
y = –(2)2 + 2(2) + 10
y = –4 + 4 + 10
y = 10

(b)



(c) From graph
(i) When x = –1.5; y = 4.6
(ii) When y = 8.2; x = 2.7

(d)
y = –x2 + 2x + 10 ……. (1)
0 = –x2 – x + 7 ………. (2)
(1) – (2): y = 3x + 3 



From graph, the values of x are –3.2 and 2.2.


4.10 SPM Practice (Long Questions)

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Question 9:
 (a) Given  1 s ( −4 2 −5 3 )( t −2 5 −4 )=( 1 0 0 1 ), find the value of s and of t.

(b) Using matrices, calculate the value of x and of y that satisfy the following matrix equation:
( −4 2 −5 3 )( x y )=( 1 2 )


Solution:
(a) 1 s ( t −2 5 −4 )= ( −4 2 −5 3 ) −1 = 1 ( −4 )( 3 )−( 2 )( −5 ) ( 3 −2 5 −4 ) = 1 −2 ( 3 −2 5 −4 ) ∴s=−2, t=3

(b) ( −4 2 −5 3 )( x y )=( 1 2 )   ( x y )= 1 −2 ( 3 −2 5 −4 )( 1 2 )   ( x y )= 1 −2 ( ( 3 )( 1 )+( −2 )( 2 ) ( 5 )( 1 )+( −4 )( 2 ) )   ( x y )= 1 −2 ( −1 −3 )   ( x y )=( 1 2 3 2 ) ∴x= 1 2 ,  y= 3 2



Question 10 (5 marks):
During the sport day, students used coupons to buy food and drink. Ali and Larry spent RM31 and RM27 respectively. Ali bought 2 food coupons and 5 drinks coupons while Larry bought 3 coupons and 1 drinks coupon.
Using the matrix method, calculate the price, in RM, of a food coupon and of a drinks coupon.

Solution:
Ali spent RM31. He bought 2 food coupons and 5 drinks coupons.
Larry spent RM27. He bought 3 food coupons and 1 drinks coupons.
x = price of one food coupon
y = price of one drinks coupon

( 2    5 3    1 )( x y )=( 31 27 )             ( x y )= 1 2( 1 )−5( 3 ) ( 1    −5 −3     2 )( 31 27 )             ( x y )= 1 2−15 ( 1( 31 )+( −5 )( 27 ) −3( 31 )+2( 27 ) )             ( x y )= 1 −13 ( −104 −39 )             ( x y )=( 8 3 ) x=8 and y=3 Thus, the price for a food coupon is RM8 and the price for drink coupon is RM3.

4.10 SPM Practice (Long Questions)

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Question 7:
(a) Find the inverse matrix of ( 3 2 5 4 ).
(b) Ethan and Rahman went to the supermarket to buy cucumbers and carrots. Ethan bought 3 cucumbers and 2 carrots for RM9. Rahman bought 5 cucumbers and 4 carrots for RM16.
By using matrix method, find the price, in RM, of a cucumber and the price of a carrot. 


Solution:
(a) Inverse matrix of ( 3 2 5 4 ) = 1 12−10 ( 4 −2 −5 3 ) = 1 2 ( 4 −2 −5 3 ) =( 2 −1 − 5 2 3 2 )

(b) 3x+2y=9.................( 1 ) 5x+4y=16...............( 2 ) ( 3 2 5 4 )( x y )=( 9 16 )               ( x y )=( 2 −1 − 5 2 3 2 )( 9 16 )               ( x y )=( ( 2 )( 9 )+( −1 )( 16 ) ( − 5 2 )( 9 )+( 3 2 )( 16 ) )               ( x y )=( 18−16 − 45 2 +24 )               ( x y )=( 2 3 2 ) x=2,  y= 3 2 ∴Price of a cucumber=RM2    Price of a carrot=RM1.50



Question 8:
The inverse matrix of   ( 4 −1 2 5 ) is t( 5 1 −2 n ).
(a) Find the value of n and of t.
(b) Write the following simultaneous linear equations as matrix equation:
4x – y = 7
2x + 5y = –2
Hence, using matrix method, calculate the value of x and of y.


Solution:
(a) t( 5 1 −2 n )= ( 4 −1 2 5 ) −1 = 1 ( 4 )( 5 )−( −1 )( 2 ) ( 5 1 −2 4 ) = 1 22 ( 5 1 −2 4 ) ∴t= 1 22 , n=4

(b) ( 4 −1 2 5 )( x y )=( 7 −2 )   ( x y )= 1 22 ( 5 1 −2 4 )( 7 −2 )   ( x y )= 1 22 ( ( 5 )( 7 )+1( −2 ) ( −2 )( 7 )+( 4 )( −2 ) )   ( x y )= 1 22 ( 35−2 −14−8 )   ( x y )= 1 22 (  33 −22 )   ( x y )=(   3 2 −1 ) ∴x= 3 2 ,  y=−1

4.10 SPM Practice (Long Questions)

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Question 5:
(a) Given  1 14 ( 2 s −4 t )( t −1 4 2 )=( 1 0 0 1 ), find the value of s and of t.
(b) Write the following simultaneous linear equations as matrix form:
3x – 2y = 5
9x + y = 1
Hence, using matrix method, calculate the value of x and y.

Solution:
(a) 1 14 ( 2 s −4 t )( t −1 4 2 )=( 1 0 0 1 ) 1 14 ( 2t+4s −2+2s −4t+4t 4+2t )=( 1 0 0 1 ) −2+2s 14 =0   2s=2      s=1 4+2t 14 =1 4+2t=14 2t=10 t=5

(b) ( 3 −2 9 1 )( x y )=( 5 1 )   ( x y )= 1 21 ( 1 2 −9 3 )( 5 1 )   ( x y )= 1 21 ( ( 1 )( 5 )+( 2 )( 1 ) ( −9 )( 5 )+( 3 )( 1 ) )   ( x y )= 1 21 ( 7 −42 )   ( x y )=( 1 3 −2 ) ∴x= 1 3 ,  y=−2



Question 6:
It is given that matrix P=( 6 −3 −5 2 ) and matrix Q= 1 m ( 2 3 5 n )  such that PQ=( 1 0 0 1 ).
(a) Find the value of m and of n.
(b) Write the following simultaneous linear equations as matrix form:
6x – 3y = –24
–5x + 2y = 18
Hence, using matrix method, calculate the value of x and y.

Solution:
(a) m=6( 2 )−( −3 )( −5 )   =12−15 m=−3 n=6

(b) ( 6 −3 −5 2 )( x y )=( −24 18 )   ( x y )= 1 12−15 ( 2 3 5 6 )( −24 18 )   ( x y )= 1 −3 ( ( 2 )( −24 )+( 3 )( 18 ) ( 5 )( −24 )+( 6 )( 18 ) )   ( x y )= 1 −3 ( 6 −12 )   ( x y )=( −2 4 ) ∴x=−2,  y=4

4.10 SPM Practice (Long Questions)

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Question 3:
It is given that Q  ( 3 2 6 5 ) = ( 1 0 0 1 ) , where Q is a 2 x 2 matrix.
(a) Find Q.
(b) Write the following simultaneous linear equations as matrix equation:
3u + 2v = 5
6u + 5v = 2
Hence, using matrix method, calculate the value of u and v.

Solution:
(a) 
Q = ( 3 2 6 5 ) − 1 Q = 1 3 ( 5 ) − 2 ( 6 ) ( 5 − 2 − 6 3 ) Q = 1 3 ( 5 − 2 − 6 3 ) Q = ( 5 3 − 2 3 − 2 1 )

(b)
( 3 2 6 5 ) ( u v ) = ( 5 2 ) ( u v ) = 1 3 ( 5 − 2 − 6 3 ) ( 5 2 ) ( u v ) = 1 3 ( ( 5 ) ( 5 ) + ( − 2 ) ( 2 ) ( − 6 ) ( 5 ) + ( 3 ) ( 2 ) ) ( u v ) = 1 3 ( 21 − 24 ) ( u v ) = ( 7 − 8 ) ∴ u = 7 , v = − 8


Question 4:
It is given that Q ( 3 − 2 5 − 4 ) = ( 1 0 0 1 )  , where Q is a 2 × 2 matrix.
(a) Find the matrix Q.
(b) Write the following simultaneous linear equations as matrix equation:
3x – 2y = 7
5x – 4y = 9
Hence, using matrix method, calculate the value of x and y.

Solution:
(a)
Q = 1 3 ( − 4 ) − ( 5 ) ( − 2 ) ( − 4 2 − 5 3 ) = − 1 2 ( − 4 2 − 5 3 ) = ( 2 − 1 5 2 − 3 2 )

(b)
( 3 − 2 5 − 4 ) ( x y ) = ( 7 9 ) ( x y ) = − 1 2 ( − 4 2 − 5 3 ) ( 7 9 ) ( x y ) = − 1 2 ( ( − 4 ) ( 7 ) + ( 2 ) ( 9 ) ( − 5 ) ( 7 ) + ( 3 ) ( 9 ) ) ( x y ) = − 1 2 ( − 10 − 8 ) ( x y ) = ( 5 4 ) ∴ x = 5 , y = 4

4.9 SPM Practice (Short Questions)

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4.9.2 Matrices, SPM Practice (Short Questions)
Question 5:
Given that ( 3   x ) ( x − 1 ) = ( 18 ) ,  find the value of x.

Solution:
( 3   x ) ( x − 1 ) = ( 18 )
[3 × x + x (–1)] = (18)
3x – x = 18
2x = 18
x = 9



Question 6:
( 3 4 − 2 3 ) ( 5 − 2 ) =

Solution:
( 3 4 − 2 3 ) ( 5 − 2 ) = ( ( 3 ) ( 5 ) + ( 4 ) ( − 2 ) ( − 2 ) ( 5 ) + ( 3 ) ( − 2 ) ) = ( 15 − 8 − 10 − 6 ) = ( 7 − 16 )



Question 7:
( 2 4 − 3 4 0 1 ) ( 1 − 3 ) =

Solution:
Order of the product of the two matrices
= ( 3 × 2 ) ( 2 × 1 ) = ( 3 × 1 )

( 2 4 − 3 4 0 1 ) ( 1 − 3 ) = ( 2 ( 1 ) + 4 ( − 3 ) ( − 3 ) ( 1 ) + 0 ( − 3 ) ( 4 ) ( 1 ) + 1 ( − 3 ) ) = ( 2 − 12 − 3 + 0 4 − 3 ) = ( − 10 − 3 1 )



Question 8:
( 1  −1   2 )( 5 1 −3 2 0 4 )=

Solution:
Order of the product of the two matrices
= ( 1 × 3 ) ( 3 × 2 ) = ( 1 × 2 )

( 1  −1   2 )( 5 1 −3 2 0 4 )=
= (1×5 + (–1)(–3) + (2)(2)  1×1 + (–1)(0) + (2)(4))
= (5 + 3 + 4   1 + 0 + 8)
= (12   9)

4.5 Multiplication of Two Matrices (Sample Question 2)

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Example 2:
Find the values of m and n in each of the following matrix equations.
( a ) ( 3 m ) ( 1       n ) = ( 3 12 − 2 − 8 ) ( b ) ( m 2 − 3 1 ) ( 2 n ) = ( 12 4 + 2 n ) ( c ) ( m − 3 − 1 1 ) ( − 1 2 4 n ) = ( − 14 − 11 5 3 )
 
Solution:
(a) ( 3 m ) ( 1   n ) = ( 3 12 − 2 − 8 ) ( 3 3 n m m n ) = ( 3 12 − 2 − 8 ) m = − 2 ,   3 n = 12 n = 4

(b) ( m 2 − 3 1 ) ( 2 n ) = ( 12 4 + 2 n ) ( 2 m + 2 n − 6 + n ) = ( 12 4 + 2 n ) − 6 + n = 4 + 2 n n = − 10 2 m + 2 n = 12 2 m + 2 ( − 10 ) = 12 2 m − 20 = 12 2 m = 32 m = 16

(c) ( m − 3 − 1 1 ) ( − 1 2 4 n ) = ( − 14 − 11 5 3 ) ( − m + ( − 12 ) 2 m + ( − 3 n ) 1 + 4 − 2 + n ) = ( − 14 − 11 5 3 ) − m − 12 = − 14 − m = − 2 m = 2 − 2 + n = 3 n = 5