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Long Questions (Question 7 & 8)

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4.2.4 Simultaneous Equations, Long Questions 
Question 7:
Solve the following simultaneous equations.
5y – 6x= 2
4 y x − 3 x y = 4.

Solution:
5y – 6x= 2 ----- (1)
4 y x − 3 x y = 4  ------- (2) From (1), y = 2 + 6 x 5

Substitute (3) into (2),
4 ( 2 + 6 x 5 ) x − 3 x ( 2 + 6 x 5 ) = 4 8 + 24 x 5 x − 15 x 2 + 6 x = 4 ( 8 + 24 x ) ( 2 + 6 x ) − ( 15 x ) ( 5 x ) 5 x ( 2 + 6 x ) = 4

16 + 48x + 48x + 144x2 – 75x2 = 20x (2 + 6x)
69x2 + 96x + 16 = 40x + 120x2
51x2 – 56x – 16 = 0
(3x – 4)(17x + 4) = 0
3x – 4 = 0 or 17x + 4 = 0
x = 4 3   or    x = − 4 17
  

Substitute  x = 4 3  into (3), y = 2 + 6 ( 4 3 ) 5 = 2

Substitute  x = − 4 17  into (3), y = 2 + 6 ( − 4 17 ) 5 = 2 17

∴ The solutions are  ( 4 3 , 2 )  and  ( − 4 17 , 2 17 ) .




Question 8:
Solve the following simultaneous equations.
x+2y=1 2 x 2 + y 2 +xy=5
Give your answer correct to three decimal places.

Solution:
x+2y=1..........( 1 ) 2 x 2 + y 2 +xy=5..........( 2 ) x=1−2y..........( 3 ) Substitute ( 3 ) into ( 2 ), 2 ( 1−2y ) 2 + y 2 +( 1−2y )y=5 2( 1−4y+4 y 2 )+ y 2 +y−2 y 2 =5 2−8y+8 y 2 − y 2 +y−5=0 7 y 2 −7y−3=0 a=7,b=−7,c=−3 x= −b± b 2 −4ac 2a y= −( −7 )± ( −7 ) 2 −4( 7 )( −3 ) 2( 7 ) y= 7± 133 14 y=1.324  or  −0.324 From x=1−2y When y=1.324,  x=1−2( 1.324 )=1.648 When y=−0.324,  x=1−2( −0.324 )=1.648 ∴The solutions are ( 1.648,1.324 ) and ( 1.648,−0.324 ).

Long Questions (Question 5 & 6)

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Question 5:

Lisa has a rectangular plot of land. She plants orchid and rears fish in the areas as shown on diagram above. The area used for planting orchid is 460 m2 and the perimeter of the rectangular fish pond is 48 m. Find the value of x and y.


Solution:
Area used for planting orchid = 460 m2
10 (30 – y) + xy = 460
300 – 10y + xy = 460
xy – 10y = 160
y (x – 10) = 160

y = 160 x − 10  ------- (1)

Perimeter of the rectangular fish pond = 48 m
2 (x – 10) + 2 (30 – y) = 48
2x – 20 + 60 – 2y = 48
2x – 2y= 8
x – y = 4
x = 4 + y ------ (2)


Substitute (2) into (1):
y = 160 x − 10 y = 160 4 + y − 10 y = 160 y − 6 y 2 − 6 y − 160 = 0 ( y − 16 ) ( y + 10 ) = 0 y = 16   or    y = − 10  (not accepted)

From (2),
When y = 16
x = 4 + 16 = 20



Question 6:

In the diagram above, PQRS is a rectangular piece of paper with an area of 112 cm2. A semicircle STR is cut from this piece of paper. The perimeter of the remaining piece of paper is 52 cm. Using  π = 22 7 , find the integer value of x and y.

Solution:
Area of the rectangle PQRS = 112 cm2
Therefore, (14x)(2y) = 112
 28xy= 112
 xy = 4 ------ (1)

Perimeter of PSTRQ = 52 cm
PS + QR + PQ + Length of arc STR = 52
2y + 2y + 14x + ½ (2πr) = 52
4y + 14x +   ( 22 7 ) ( 7 x ) = 52
4y + 14x + 22x = 52
4y + 36x = 52
y + 9x = 13 ------ (2)

From equation (2) : y = 13 – 9x ------ (3)

Substitute (3) into (1) :
x (13 – 9x) = 4
13x – 9x2 = 4
9x2 – 13x + 4 = 0
(x – 1)(9x – 4) = 0
x = 1   or   4 9  (not an interger, therefore not accepted)

From (3) :
When x = 1,
y = 13 – 9(1) = 4. 

Long Questions (Question 3 & 4)

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Question 3:
Solve the following simultaneous equations.
3y – 2x= – 4
y2 + 4x2 = 2

Solution:
3y – 2x= – 4 -----(1)
y2 + 4x2= 2 -----(2)

From (1), y = 2 x − 4 3  ------- ( 3 ) Substitute (3) into (2), ( 2 x − 4 3 ) 2 + 4 x 2 = 2 ( 4 x 2 − 16 x + 16 9 ) + 4 x 2 = 2 4 x 2 − 16 x + 16 + 36 x 2 = 18     ( × 9 ) 40 x 2 − 16 x − 2 = 0 20 x 2 − 8 x − 1 = 0 ( 10 x + 1 ) ( 2 x − 1 ) = 0 x = − 1 10   or   x = 1 2 Substitute the values of  x  into (3), When  x = − 1 10 , y = 2 ( − 1 10 ) − 4 3 = − 1 2 5 When  x = 1 2 , y = 2 ( 1 2 ) − 4 3 = − 3 3 = 1 The solutions are  x = − 1 10 ,   y = − 1 2 5  and  x = 1 2 ,   y = 1.





Question 4:
Solve the simultaneous equations x – 3y = –1 and y + yx – 2x = 0.
Give your answers correct to three decimal places.

Solution:
x – 3y = –1 -----(1)
y + yx – 2x = 0 -----(2)
From (1),
x = 3y – 1 -----(3)
Substitute (3) into (2),
y + y (3y – 1) – 2(3y – 1)  = 0
y + 3y2 – y – 6y+ 2 = 0
3y2 – 6y + 2 = 0

a = 3 ,   b = − 6 ,  c = 2 y = − b ± b 2 − 4 a c 2 a y = − ( − 6 ) ± ( − 6 ) 2 − 4 ( 3 ) ( 2 ) 2 ( 3 ) y = 6 ± 12 6 y = 1.577  or 0 .423

Substitute the values of y into (3).
When y = 1.577,
x = 3 (1.577) – 1 = 3.731 (correct to 3 decimal places)

When y = 0.423,
x = 3 (0.423) – 1 = 0.269 (correct to 3 decimal places)

The solutions are x = 3.731, y = 1.577 and x = 0.269, y = 0.423.


Long Questions (Question 1 & 2)

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Question 1:
Solve the following simultaneous equations.
y + 2 x = 2 2 x + 1 y = 5

 Solution:
y+2x=2−−−−(1) 2 x + 1 y =5−−−−(2) y=2−2x−−−−(3) substitute (3) into (2), 2 x + 1 2−2x =5 2( 2−2x )+x x( 2−2x ) =5 4−4x+x=5x( 2−2x ) 4−3x=10x−10 x 2 10 x 2 −13x+4=0 ( 5x−4 )( 2x−1 )=0 5x−4=0     or     2x−1=0 x= 4 5            or     x= 1 2 Substitute values of x into (3), When x= 4 5  ,  y=2−2( 4 5 )= 2 5 When x= 1 2 y=2−2( 1 2 )=1 The solutions are x= 4 5 , y= 2 5  and x= 1 2 , y=1




Question 2:
Solve the following simultaneous equations.
x – 3y + 5 = 3y + 5y2– 6 – x = 0

Solution:
x – 3y + 5 = 0
x = 3y – 5 -----(1)
3y + 5y2 – 6 – x = 0 -----(2)

Substitute (1) into (2),
3y + 5y2 – 6 – (3y – 5) = 0
3y + 5y2 – 6 – 3y + 5 = 0
5y2 – 1 = 0
5y2  = 1
y = ±0.447

Substitute the values of y into (1),
When y = 0.447
x = 3 (0.447) – 5
x = –3.659

When y = – 0.447
x = 3 (–0.447) – 5
x = –6.341

The solutions are x = –3.659, y = 0.447 and x = –6.341, y = – 0.447.

Simultaneous Equations (Example 1 & 2)

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Example 1:
Solve the simultaneous equations.
x + 1 4 y = 1  and  y 2 − 8 = 4 x .

Solution:
x + 1 4 y = 1 → ( 1 ) y 2 − 8 = 4 x → ( 2 ) x = 1 − 1 4 y → ( 3 )

Substitute (3) into (2),
y 2 − 8 = 4 ( 1 − 1 4 y ) y 2 − 8 = 4 − 4 4 y
y2 + y – 12 = 0
(y + 4)(y – 3) = 0
y = –4 or y = 3 

Substitute the values of y into (3),
when y=−4,  x=1− 1 4 (−4)=2 when y=3,  x=1− 1 4 (3)= 1 4 The solutions are x=2, y=−4 and x= 1 4 , y=3


Example 2:
Solve the simultaneous equations 2x + y = 1 and 2x2+ y2 + xy = 5.
Correct your answer to three decimal places.

Solution:
2x + y = 1-----(1)
2x2 + y2+ xy = 5-----(2)

From (1),
y = 1 – 2x-----(3)

Substitute (3) into (2).
2x2 + (1 – 2x)2 + x(1 – 2x) = 5
2x2 + (1 – 2x)(1 – 2x) + x – 2x2 = 5
1 – 2x – 2x + 4x2 + x – 5 = 0
4x2 – 3x – 4 = 0

From  x = − b ± b 2 − 4 a c 2 a a = 4 ,   b = − 3 ,  c = − 4 x = − ( − 3 ) ± ( − 3 ) 2 − 4 ( 4 ) ( − 4 ) 2 ( 4 ) x = 3 ± 73 8 x = − 0.693  or  1.443

Substitute the values of x into (3).
When x = –0.693,
y = 1 – 2 (–0.693) = 2.386 (correct to 3 decimal places)

When x = 1.443,
y = 1 – 2 (1.443) = –1.886 (correct to 3 decimal places)

The solutions are x = –0.693, y  = 2.386 and x = 1.443, y = –1.886.

Quadratic Functions, SPM Practice (Long Questions)

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Question 5:


Diagram above shows the graphs of the curves y = x2 + x – kx + 5 and y = 2(x – 3) – 4h that intersect the x-axis at two points. Find
(a) the value of k and of h,
(b) the minimum value of each curve.


Solution:
(a)
y= x 2 +x−kx+5 = x 2 +( 1−k )x+5 = [ x+ ( 1−k ) 2 ] 2 − ( 1−k 2 ) 2 +5 axis of symmetry of the graph is x=− ( 1−k ) 2

y=2 ( x−3 ) 2 −4h axis of symmetry of the graph is x=3. ∴ − 1−k 2 =3 −1+k=6 k=7

Substitute k=7 into equation y= x 2 +x−7x+5   = x 2 −6x+5 At x-axis,y=0; x 2 −6x+5=0 ( x−1 )( x−5 )=0 x=1,5

At point ( 1,0 ) Substitute x=1,y=0 into the graph: y=2 ( x−3 ) 2 −4h 0=2 ( 1−3 ) 2 −4h 4h=2( 4 ) 4h=8 h=2

(b)
For y= x 2 −6x+5 = ( x−3 ) 2 −9+5 = ( x−3 ) 2 −4 ∴ Minimum value is −4. For y=2 ( x−3 ) 2 −8, minimum value is−8.

Quadratic Functions, SPM Practice (Long Questions)

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3.9.1 Quadratic Functions, SPM Practice (Long Questions)

Question 1:
Without drawing graph or using method of differentiation, find the maximum or minimum value of the function y = 2 + 4x – 3x2. Hence, find the equation of the axis of symmetry of the graph.

Solution:
By completing the square for the function in the form of y = a(x + p)2+ q to find the maximum or minimum value of the function.

y = 2 + 4x – 3x2
y = – 3x2 + 4x + 2 ← (in general form)
y = − 3 [ x 2 − 4 3 x − 2 3 ] y = − 3 [ x 2 − 4 3 x + ( − 4 3 × 1 2 ) 2 − ( − 4 3 × 1 2 ) 2 − 2 3 ] y = − 3 [ ( x − 2 3 ) 2 − ( − 2 3 ) 2 − 2 3 ]  

y = − 3 [ ( x − 2 3 ) 2 − 4 9 − 6 9 ] y = − 3 [ ( x − 2 3 ) 2 − 10 9 ] y = − 3 ( x − 2 3 ) 2 + 10 3 ← in the form of  a ( x + p ) 2 + q

Since a = –3 < 0,
Therefore, the function y has a maximum value of 10 3 . 
x − 2 3 = 0 x = 2 3
Equation of the axis of symmetry of the graph is x = 2 3 .   



Question 2:

The diagram above shows the graph of a quadratic function y = f(x). The straight line y = –4 is  tangent to the curve y = f(x).
(a) Write the equation of the axis of symmetry of the function f(x).
(b)   Express f(x) in the form of (x + p)2 + q , where p and q are constant.
(c) Find the range of values of x so that
(i) f(x) < 0, (ii) f(x) ≥ 0.

Solution:
(a)
x-coordinate of the minimum point is the midpoint of (–2, 0) and (6, 0)
= = − 2 + 6 2 = 2  
Therefore, equation of the axis of symmetry of the function f(x) is x = 2.



(b)
Substitute x = 2 into x + p = 0,
2 + p = 0
p = –2
and q = –4 (the smallest value of f(x))
Therefore, f(x) = (x + p)2 + q
f(x) = (x – 2)2 – 4

(c)(i) From the graph, for f(x) < 0, range of values of x are –2 < x < 6 ← (below x-axis).

(c)(ii) From the graph, for f(x) ≥ 0, range of values of x are x ≤ –2 or x ≥ 6 ← (above x-axis).


Quadratic Functions, SPM Practice (Short Questions)

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Question 5:
Find the range of values of k if the quadratic equation 3(x2 – kx – 1) = k – k2 has two real and distinc roots.

Solution:
3( x 2 −kx−1 )=k− k 2 3 x 2 −3kx−3−k+ k 2 =0 3 x 2 −3kx+ k 2 −k−3=0 a=3,b=−3k,c= k 2 −k−3 In cases of two real and distinc roots, b 2 −4ac>0 is applied. ( −3k ) 2 −4( 3 )( k 2 −k−3 )>0 9 k 2 −12 k 2 +12k+36>0 −3 k 2 +12k+36>0 − k 2 +4k+12>0 k 2 −4k−12<0 ( k+2 )( k−6 )<0 k=−2,6



The range of values of k is −2<k<6.



Question 6:
Given that the quadratic equation hx2 – (h + 2)x – (h – 4) = 0 has real and distinc roots.  Find the range of values of h.

Solution:
The quadratic equation h x 2 −( h+2 )x−( h−4 )=0 has real and distinc roots. b 2 −4ac>0 is applied. ( −h−2 ) 2 −4( h )( −h+4 )>0 h 2 +4h+4+4 h 2 −16h>0 5 h 2 −12h+4>0 ( 5h−2 )( h−2 )>0 The coefficient of  h 2  is positive,  the region above the x-axis should be shaded. ( 5h−2 )( h−2 )=0 h= 2 5 ,2



The range of values of h for ( 5h−2 )( h−2 )>0 is h< 2 5  or h>2.

3.5b Nature of the Roots (Combination of Straight Line and a Curve)

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Example 2 (Combination of Straight Line and a Curve)
The straight line y = 2 k + 1   intersects the curve y = x + k 2 x   at two distinct points.  Find the range of values of k.



Example 3 (Straight Line does not intersect the curve)
Find the range of values of m for which the straight line y = m x + 6   does not meet the curve 2 x 2 − x y = 3  .