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Friday 15 July 2011

Maths - Quardiatic Equations for Competition

Solving Quadratic Equations by Factorisation

Illustrative Examples

Example;Solve for x: (x +3)(x -3) = 40.

Solution:(x +3)(x -3) = 40 => x² -9 = 40

=> x² -49 = 0 => (x -7)(x +7) = 0,        => x -7 = 0 or x +7 = 0 => x = 7 or x = -7.
Hence the roots of the given equation are 7, -7.

Example:Solve(x -3)/(x +3) + (x +3)/(x -3) = 5/2

Solution:Given (x -3)/(x +3) + (x +3)/(x -3) = 5/2

To clear the fractions, multiply both sides by L.C.M. of fractions i.e. by 2(x +3)(x -3) to get
2(x -3)² +2(x +3)² = 5(x -3)(x +3),           => 2(x² -6x +9) +2(x² +6x +9) = 5(x² -9)
=> 2x² -12x +18 +2x² +12x +18 -5x² +45 = 0
=> -x² +81 = 0 => x² -81 = 0 => (x -9)(x +9) = 0
=> x -9 = 0 or x +9 = 0
Hence the values x = 9, x = -9 satisfy the given equation.

Example:Three consecutive natural numbers are such that the square of the middle number exceeds the difference of the squares of the other two by 60. Assume the middle number to be x and form a quadratic equation satisfying the above statement. Hence find the three numbers.

Solution:Since the middle number of the three consecutive numbers is x, the other two numbers are x -1 and x +1.

According to the given condition, we have
    x² = [(x +1)² -(x -1)²] +60
=> x² = (x² +2x +1) -(x² -2x +1) +60 = 4x +60
=> x² -4x -60 = 0 => x
² - (10-6)x -60=0, x² - 10x +6x -60=0, 
x(x-10)+6(x-10)=0,  (x -10)(x +6) = 0, => x = 10 or -6
Since x is a natural number, we get x = 10
Hence the three numbers are 9, 10, 11

                                       Excercise:
 
Q1. x² - 5 x = 0 a. 5 b. 7 c. 9 d. 11
Q2. x (2x +1) = 6 a. -2, 3/2 b. ½, 3/2 c. 1, 2 d. 2, 2
Q3. x² - 3x -10 = 0 a. 5, 2 b. 2, 4 c. 5, 4 d. 5, -2
Q4. 3x² - 5x - 12 = 0 a. 3, -4/3 b. 3, 4 c. 2, 3 d. 3/5, 5/3
Q5. 3x² = x + 4 a. -1, -3 b. -1, 4/3 c. 4/3, 1 d. 1, 1/2
Q6. (x + 2)(x - 3) = 6 a. -3, 4 b. 3, -4 c. 4, 4 d. 3, 3
Q7. 3x – 8 / x = 2 a. 2, -4/3 b. 2, 4/3 c, 1, 2 d, 2, 3
Q8. (x + 2) / (x +3) = (2 x -3) / (3 x - 7) a. -1, 5 b. 1, -5 c. 1, 1 d. 5, 5
Q9. 8/(x +3) - 3/(2 -x) = 2 a. 5, -1/2 b. 5, ½ c. 5, 5 d, 1, 1
Q10. J and K has total 45 balls. If both lost 5-5 balls, than the multiplication of the balls remaining with them is 124. How much they have initially?
          a. 36, 9 b. 36, 36 c. 9, 9 c. Cant find e. None
Q11. The area of a ractangle fig is 506 sqmt. If length is one more than the width. Find the length.
          a. 22, 23 b. 23, 24 c. 24, 25 c. Cant Find e. None
Q12. Rohan's mother is 26 year older than him. The product of their ages after 3 years will be 360. Find Rohan's current age.
          a. 5 b. 7 c. 9 c. 11
Q13. The diagonal of a ractangle field is 60 m more than the smaller side. If longer side is 30 m more than the shorter side. Find the shorter side.
            a. 90 b. 150 c. 120 c. Cant Find e. None
Q14. The differance of the squares of two numbers is 180. The sq of smaller no is eight times the big no. Find the smaller one.
          a. 11 b. 12 c. 13 d. 14
Q15. The sum of the area of two squares is 468 sqm. If the diffrence of their perimeter is 24m. Find the side of smaller area.
         a. 10 b. 11 c. 12 d. 13
Q16. The perimeter of a ractangle field is 82 m and its area is 400 sq m. Find the breadth of the field is:
          a.14 m b. 16 m c. 18 m d. 20 m

Answers: 1a, 2a, 3d, 4a, 5b, 6a, 7a, 8a,9a, 10a, 11a, 12b, 13a, 14b, 15c, 16b

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