Quadratic expressions and equation 2 Questions and Answers - Form 3 Topical Mathematics

Questions

1.  
   1. Use a convenient scale to draw the graph of y = -x² + 5x – 3 for the range -2 ≤ x ≤ 6
   2. Use your graph to determine the roots of the equation 5x – x² – 3 = 0
   3. Use your graph to solve the equation 2x – x² + 3 = 0 by drawing a suitable straight line
2. Find a quadratic equation whose roots are 2.5 + √3 and 2.5 – √3, expressing it in the form ax² + bx + c = 0 Where a, b and c are integers.
3.  
   1. Complete the table below for the equation y = x² + 3x – 6 for -6 ≤ x ≤ 4
      

      x	-6	-5	-4	-3	-2	-1	0	1	2	3	4
      y	12			-6			-6				22

   2. Using a scale 1cm to represent 2 units in both axes. Draw the graph of y = x² + 3x – 6
   3. Use your graph to solve:-
      1. X² + 3X = 6
      2. X² + 3X – 2 = 0
4.  
   1. Complete the table for the function: y = 2x² + 3x + 1
      

      x	-4	-3	-2	-1	0	1	2	3
      2x2		18			0			18
      3x+1		-7			0			10
      y		10			1	6

   2. Use the table in (a) above to draw the graph : y = 2x² + 3x + 1 for -4 ≤ x ≤ 3
   3. Use the graph in (b) to solve the equation :-
      1. 2x² + 4x – 3 = 0
      2. x² + ³/₂x + 2 = 3
5. A youth group decided to raise Ksh 480,000 to buy a piece of land costing Ksh. 80,000 per hectare. Before the actual payment was made, four of the members pulled out and each of those remaining had to pay an additional Kshs. 20,000.
   1. If the original number of the group members was x, write down;
      1. An expression of how much each was to contribute originally.
      2. An expression of how the remaining members were to contribute after the four pulled out.
   2. Determine the number of members who actually contributed towards the purchase of the land.
   3. Calculate the ratio of the supposed original contribution to the new contribution.
   4. If the land was sub-divided equally, find the size of land each member got. (2 mk)
6.  
   1. Draw the graph of y = 2x² + x – 2 given the range -3 ≤ x ≤ 2
   2. Use your graph above to solve
      1. 2x² + x – 2 =0
      2. 2x² + x – 3 =0
      3. 2x² + x-5 =0
7.  
   1. Use trapezoidal rule to find the area between the curve y = x² + 4x + 4, the x- axis and the co-ordinates x = -2 and x = 1. Take values of x at intervals of ½ unit.
   2. Use integration to find the exact area. Hence find the percentage error in your approximation.
8.  
   1. Use trapezoidal rule to find the area between the curve y = x² + 4x + 4, the x- axis and the co-ordinates x = -2 and x = 1. Take values of x at intervals of ½ unit.
   2. Use integration to find the exact area. Hence find the percentage error in your approximation.
9. Draw the graph of y = 2x² – 4x - 5 for x between -3 and 5 on the grid provided.
   1. State the line of symmetry for the graph
   2. State the range of values for which 2x² – 4x – 5 ≤ 0
   3. On the same set of axes, draw the graph of y=2x +3
   4. Determine the solutions to the equation: 2x² – 4x – 5 = 2x +3
10. Complete the table below for the equation y = 5 + 3x -2x² by filling in the blank spaces.
    

    X	-2	-1.5	-1	-0.5	0	0.5	1	1.5	2	2.5	3	3.5
    Y	-9			3		6	6				-4

    
    
    1. Use the values from the table above to draw the graph of y = 5 + 3x -2x² (3mks)
    2. Use the graph to:-
       1. Find the maximum point of the function 5 + 3x -2x²
       2. Determine the range of values and give the integral values which satisfy
          the inequality 5 + 3x -2x² ≥ -2
11.  
    1. Complete the table below for the function y = 2x² + 4x – 3
       

       x	-4	-3	-2	-1	0	1	2
       2x2	32		8	2	0
       4x-3			-11		-3		5
       y			-3			3	13

    2. Draw the graph of the function y = 2x² + 4x – 3 and use your graph to estimate the roots of the equation 2x² + 4x – 3 = 0.
    3. In order to solve graphically the equation 2x² + x – 5 = 0, a straight line must be drawn to intersect the curve y = 2x² + 4x – 3. Determine the equation of this line, draw it and hence obtain the roots of the equation 2x² + x – 5 = 0 to 1 decimal place.
12.  
    1. Complete the table for the function y = 1 – 2x – 3x² for - 3 ≤ x ≤ 3.
       

       x	-3	-2	-1	0	1	2	3
       -3x2	-27		-3	0		-12
       -2x		4		0			-6
       1	1	1	1	1	1	1	1
       y	-20			1		-15

    2. Using the table above, draw the graph of y = 1 – 2x – 3x² (Scale 1 cm represent 0.5 units on
       x-axis and 1 cm rep 2 units on the y – axis on the grid provided.
    3. Use the graph in (b) above to solve.
       1. 1 – 2x – 3x² = 0
       2. 2 – 5x – 3x² = 0
13. A quadratic equation x² + ax – b = 0 has roots 1 and -5 , determine the values of a and b
14. Find a quadratic equation whose roots are 1.5 + √2 and 1.5 – √2 , expressing it in the form ax² + bx + c = 0, where a, b, and c are integers
15. If a² + b² = 89 and a + b = 13
    1. Find the values of;
       1. a² + 2ab + b²
       2. 2ab
       3. a² – 2ab + b²
       4. a - b
    2. Determine the values of a and b
16. 1. Complete the table below for the function y = 2x³ +5x² - x - 6 (2 mks)
       

       x	-4	-3	-2	-1	0	1	2
       2x3	-128	-54			0	2	16
       5x2	80	45	20	5	0	5	20
       -x	4	3			0	-1
       -6	-6	-6	-6	-6	-6	-6	-6
       y	-50				-6	0

    2. On the grid provided draw the graph y = 2x³ +5x² - x - 6 for -4 ≤x ≤2. Use 2cm to represent 1 unit on the x-axis and 1 cm to represent 5 units on the y – axis (4 mks)
    3. By drawing a suitable line, use the graph in (b) to solve the
       1. 2x³ + 5x² + x - 4 = 0
       2. 2x³ + 5x² - x + 2 = 0 

Answers

1.  
   1.  
      

      x	-2	-1	0	1	2	3	4	5	6
      y	-17	-9	-3	1	3	3	1	-3	-9

   2. y = 5x – x² - 3
      0 = 5x –x² -3
      y = 0
      x = 0.75 or 4.3 ± 0.1
   3. y = 5x – x² - 3
      0 = 2x – x² + 3
      y = 3x – 6
      
      x 0 -1 2
      y -6 -9 0
      
      x = -1 or 3 + 0.1
      
      
2. x – 2.5 - √3   x – 2.5 + √3 = 0
   x² – 2.5x + x√3 - 2.5x + 6.25 – 2.5 √3
   x√3 + 2.5 √3 = 0
   x² – 5x + 6.25 – 3 = 0
3.  
   1.  
      

      x	-6	-5	-4	-3	-2	-1	0	1	2	3	4
      y		04	-2		-8	-8		-2	4	12

   2.  
   3.  
      1. x² + 3x – 6 = 0
         x= - 4.5 or 1.5 ± 0.2
      2. y = x² + 3x -6
         x² + 3x -2
         y = -4
4.  
   1.  
      

      x	-4	-3	-2	-1	0	1	2	3
      y	21	10	3	0	1	6	15	28

   2.  
      
   3.  
      1. 2x² + 3x + 1 = 0
         2x² + 4x – 3 = 0
         -x + 2 = y
         x = 0.6 or x = -2.6 ± 0.1
      2. x = 0.30 –x = -1.8 ± 0.1
5.  
   1.  
      1. ^480,000/=/_x
      2. ⁴⁸⁰⁰⁰/_(x-4)
   2. 480,000 = 480,000 + 20,000
      x – 4            x
      Multiply all hr’ by L.C.M.
      480,000x = 480,000(x – 4) + 20,000(x² – 4x)
      Dividing by 10,000
      
      48x = 48x – 192 + 2x² – 4x
      48x – 48x + 4x – 2x² + 192 = 0
      4x – 2x² + 192 = 0
      
      
      But x cannot be –ve hence
      x = ^-43.3954/₄ = 10.8489
      = 11
   3. Original : new cont.
      480,000 : 480, 000
         11             7
   4. Size of land bought = 6 hectares
      ⁶/₇ = 0.857143
      ≃ 0.8571 hectares
6.  
   
   
   
7. 
    
   
   
8.  
   
9.  
   
   
   
10.  
    1.  
       

       x	-4	-3	-2	-1	0	1	2
       2x2	32	18	8	2	0	2	8
       4x-3	-19	-15	-11	-7	-3	1	5
       y	13	3	-3	-5	-3	3	13

    2. Roots for x = -2.6 ± 0.1
       x = 0.6 ± 0.1
       y = 2x² + 4x – 3
       0 = 2x² + x – 5
       y = 3x + 2
       
       Roots read from the 2 pts of intersection of the line and curve.
       X = - 1.9 + 0.1
       X = 1.4 + 0.1
11.  
    1. 
       

       x	-3	-2	-1	0	1	2	3
       -3x2	-27	-12	-3	0	-3	-12	-27
       -2x	6	4	2	0	-2	-4	-6
       1	1	1	1	1	1	1	1
       y	-20	-7	0	1	-4	-15	-32

    2. 
        
       
       1 – 2x – 3x² = 0
       x = -1
       or x = 0.7
       
       y = -3x² – 2x + 1
       0 = -3x² – 5x + 2
       y = 0 + 3x – 1
       
       x 0 2
       y -1 5
12. x² + ax – b = 0
    (x-1) (x +5) = x² + ax – b
    x² + 4x – 5 = x² + ax –b
    a = 4, b = 5
13. Let a = 1.5 + √2
    b = 1.5 - √2
    (x – a) (x – b) = 0
    x² – xb – ax + ab = 0
    x² – x (1.5 - √2) – x (1.5 + √2) + ab = 0
    x² –1.5x +x√2) – x 1.5x -√2) = 0
    x² –3x + ab
    x² –3x + (1.5 + √2) (1.5 - √2) = 0
    x² –3x + 2.25 - 2 = 0
    x² –3x + ¼ = 0
    4x² – 12x + 1 = 0
14.  
    1.  
       1. a² + b² = 89 a + b = 13
          a² + 2ab + b² = (a + b)² = 13² = 169
       2. 2ab = 169 – 89
          = 80
       3. a² – 2ab + b² = a² + b² – 2ab
          = 89 – 80 = 9
       4. (a – b)² = 9
          a-b = ±3
    2. a + b = 13
       a – b = 3
       2a = 16