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Form the quadratic equation, whose roots are − 3 and 4.

1. Form the quadratic equation, whose roots are − 3 and 4. Roots x = – 3 and x = 4. x + 3 = 0 and x – 4 = 0. ( x + 3)( x – 4) = 0. x ( x – 4) + 3( x – 4) = 0. x 2 – 4 x + 3 x – 12 = 0. x 2 – x – 12 = 0. 2. Solve the following quadratic equations:.

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Form the quadratic equation, whose roots are − 3 and 4.

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  1. 1. Form the quadratic equation, whose roots are − 3 and 4. Roots x = – 3 and x = 4 x + 3 = 0 and x – 4 = 0 (x + 3)(x – 4) = 0 x(x – 4) + 3(x – 4) = 0 x2 – 4x + 3x – 12 = 0 x2 – x – 12 = 0

  2. 2. Solve the following quadratic equations: 2x2 – 7x – 15 = 0 (i) Method 1: 2x2 – 7x – 15 = 0 (2x + 3)(x – 5) = 0 2x + 3 = 0 x – 5 = 0 2x = – 3 x = 5 x = – 1·5

  3. 2. Solve the following quadratic equations: 2x2 – 7x – 15 = 0 (i) Method 2: 2x2 – 7x – 15 = 0 a = 2, b = – 7, c = – 15

  4. 2. Solve the following quadratic equations: 6x2 = 1 – x (ii) 6x2 = 1 – x 6x2 + x – 1 = 0 (3x – 1)(2x + 1) = 0 3x – 1 = 0 and 2x + 1 = 0 3x = 1 2x = –1

  5. 3. Solve k(k + 6) = 16 and hence find the two possible values of 3k2 + 2k − 1. k(k + 6) = 16 k2 + 6k = 16 k2 + 6k – 16 = 0 (k + 8)(k – 2) = 0 k + 8 = 0 and k – 2 = 0 k = – 8 and k = 2

  6. 3. Solve k(k + 6) = 16 and hence find the two possible values of 3k2 + 2k − 1. 3k2 + 2k – 1 k = –8 k = 2 3(–8)2 + 2(–8) – 1 3(2)2 + 2(2) – 1 3(64) + (–16) – 1 3(4) + 4 – 1 192 – 16 – 1 12 + 4 – 1 175 15

  7. 4. Solve

  8. 5. Solve the following simultaneous equations x + 3 = 2y and xy− 7y + 8 = 0. Rearrange the linear equation: x + 3 = 2y x = 2y – 3 Substitute x = 2y – 3 into the quadratic equation: xy – 7y + 8 = 0  (2y – 3)y – 7y + 8 = 0 2y2 – 3y – 7y + 8 = 0 2y2 – 10y + 8 = 0 y2 – 5y + 4 = 0 (y – 1) (y – 4) = 0 y – 1 = 0 and y – 4 = 0 y = 1 and y = 4

  9. 5. Solve the following simultaneous equations x + 3 = 2y and xy− 7y + 8 = 0. x = 2y – 3 y = 1 x = 2(1) – 3 = 2 – 3 = – 1  (–1, 1) y = 4 x = 2(4) – 3 = 8 – 3 = 5  (5, 4)

  10. 6. When a number x is subtracted from its square, the result is 42. Write down an equation in xto represent this information and solve it to calculate two possible values for x. x2 – x = 42 x2 – x – 42 = 0 (x – 7)(x + 6) = 0 x – 7 = 0 and x + 6 = 0 x = 7 and x = – 6

  11. 7. Solve the equation x2 − 6x − 18 = 0, giving your answer in the form , where

  12. 8. A ball rolls down a slope and travels a distance metres in tseconds. Find the distance, d, when the time is 3 seconds. (i)

  13. 8. A ball rolls down a slope and travels a distance metres in tseconds. Find the time taken for the ball to travel 17 m.Give your answer correct to two decimal places. (ii) Time cannot be negative, so t = –14·37 is rejected. Time = 2·37 s to travel 17 m

  14. 9. A man jogs at an average speed of xkm/hr for (x– 5) hours.If the man jogged a distance of 24 km, use the information to form an equation. (i) 24 = x(x – 5) 24 = x2 – 5x 0 = x2 – 5x – 24

  15. 9. A man jogs at an average speed of xkm/hr for (x– 5) hours.If the man jogged a distance of 24 km, Hence, solve the equation to find x. (ii) 0 = x2 – 5x – 24 0 = (x – 8)(x + 3) x – 8 = 0 and x + 3 = 0 x = 8 and x = –3 Speed cannot be negative Therefore, x = 8

  16. 10. A quadratic function has a positive coefficient on the x2 term and roots of − 5 and 0. Draw a sketch of this function. Roots of x = –5 and x = 0, means that the graph crosses the x-axis at these points. A positive coefficient on the x2 term means that the graph is U shaped.

  17. 11. Find the roots of the function f (x) = 3x2 + 4x – 6. Give your answers to two decimal places.

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