Quadratic Equation Important questions

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Important questions

Example 1. Solve: x² - 4 = 0

Solved:- x² - 4 = 0

(x + 2) (x - 2) = 0

Either x + 2 = 0 which gives x = -2

Or, x - 2 = 0, which gives x = 2

 x = -2, 2

Example 2. Solve: 8x² - 22x - 21 = 0

Solution:- 8x² - 22x - 21 = 0

8x² - (28 - 6) x - 21 = 0

8x² - 28x + 6x -21 = 0

4x(2x - 7) + 3(2x - 7) = 0

(2x - 7) (4x + 3) = 0

Either 2x - 7 = 0 or, 4x + 3 = 0

Either 2x = 7 or, 4x = -3

Either x = 7/2 or, x = -3/4

Example 3. Solve: x² + 2x - 8 = 0

Solution:- x² + 2x - 8 = 0

Here a = 1, b = 2, c = -8

 Roots are real.

Example 4. Solve: 

Solution:

1/(x + 1) + 1/(x + 2) = 1/(x + 4)

Or, (x + 2) +( x + 1) / (x+ 1) (x + 2) = 1/(x + 4)

(2x + 3)/x² + x + 2x + 2 = 1/(x + 4)

Or,  (2x + 3) (x + 4) = x² + 3x + 2

Or,  2x² + 8x + 3x + 12 = x² + 3x + 2

Or,   x² + 8x + 10 = 0

Here a = 1, b = 6, c = 8

D = b² - 4ac = 8 ² - 4 X 1 X 10

= 64 - 40

= 24

Example 5. Find the value of p for which the quadratic equation 2x² + 3x + p = 0 has real roots.

Solution:- 2x² + 3x + p = 0

a = 2, b = 3, c = p

D = b² - 4ac = 3² - 4 X 2 X P

= 9 - 8p

For real roots, 

or, 

or, 

or, 

Example 6. Find the value of   such that the quadratic equation   has equal roots.

Solution:-

Here 

D = b² - 4ac

For equal roots, 

i.e. 
Or, 

But for   the quadratic equation does not exists.

Example 8. A plane left 30 minutes later than the scheduled time and in order to reach the destination 1500km away in time, it has to increase the speed by 250 km/hr from the usual speed. Find the usual speed.

Solution:- Let the usual speed of plane be x km/hr.

The increased speed of the plane = (x + 250) km/hr.

Distance = 1500 km

or, 

or, x² + 250x - 750000 = 0

or, x² + 1000x - 750x - 750000 = 0

or, x(x + 1000) -750 (x + 1000) = 0

or, ( x + 1000) ( x - 750) = 0

or, x = -1000, 750

As speed can not be negative

 Usual speed = 750 km/hr.

Example 9. If the list price of a book is reduced by Rs. 5, a person can buy 5 more books for Rs.300. Find the original list price of the book.

Soltution:- Let the original price of the book be Rs. x.

Then the reduced price of the book = Rs. ( x - 5)

Total amount = Rs. 300

According to the question

Or, 

Or, 

Or, 5(x² - 5x) = 1500

Or, x² - 5x = 300

Or, x² - 5x - 300 = 0

Or, x² - 20x + 15x - 300 = 0

Or, x(x - 20) + 15 (x - 20) = 0

Or, (x - 20) ( x + 15) = 0

x = -15, 20

As price can not be negative x = Rs. 20.

 Example 7. A rect angular field is 16m long and 10m wide, there is a path of uniform width all around it having an area of 120sq.m final the width of the path.

Solution:- Let width of the path be xm.

Area of the outer rectangle

= (16 + 2x) (10 + 2x)

= 160 + 32x + 20x + 4x²

= 160 + 52x + 4x2

Area of the inner rectangle

= 16m X 10m

= 160m²

Area of the path

= 160 + 52x + 4x² - 160 = 120

or, 4x² + 52x - 120 = 0

or, x² + 13x - 30 = 0

or, (x + 15) (x - 2) = 0

or, x = -15, 2

As width can not be negative   x = 2 m

i. e. width of the path = 2 m

Example 11. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2? If so, find its length and breadth.

Solution:- Let breadth = x, then length = 2x

According to question, l × b = 2x × x = 800

Or,     2x² = 800

Or,       x² = 400

Or,           x = 20s

Hence, it is possible to design of length  = 2×20 = 40 m and breadth = 20 m.

Example – 12. Is the following situation possible? If so, determine their present ages.

The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.

Solution :- Let present age of one friend be x and that of the other 20 – x.

4 years ago age of first friend = x – 4, and that of the other = 20 – x – 4 = 16 – x.

According to question,  (x – 4 )(16 – x ) = 48

Or,     16x – x² – 64 + 4x =48

Or,         – x² + 20x – 112 = 0

Or,            x² – 20x + 112 = 0

Here, a = 1, b = – 20, c = 112

D = b² – 4ac = (– 20 )² – 4 ×1 × 112 = – 48 < 0.

Hence, the given situation is not possible.

Example – 13. Is it possible to design a rectangular park of perimeter 80 m and area 400 m²? If so, find its length and breadth.

Solution :- 2(l + b ) = 80

Or,    l + b = 40

Let  l = x, then b = 40 – x

A/Q    x(40 – x ) = 400

Or,    40x – x² = 400

Or,    x² – 40x + 400 = 0

Here, a = 1, b = – 40, c = 400 ;

D = b² – 4 ac = (– 40 )² – 4 × 1 × 400

= 1600 – 1600

= 0 (positive).

Hence, it is possible.

By completing square, (x – 20 )² = 0

Either,            x – 20 = 0  or, x - 20 = 0

Or,                         x = 20, 20

Thus l = 20 m and b = 20 m.

Example – 14. The sum of the reciprocals of Rehman’s ages (in years), 3 years ago and 5 years from now is 1/3. Find his present age.

Solution :- Let his present age be x 

His age 3 years ago = x – 3 and 5 years hence = x + 5

A/Q           1/(x – 3 ) + 1/(x + 5) = 1/3 
Or, (x + 5 + x – 3)/(x – 3)(x + 5) = 1/3 
Or,   (2x + 2)/(x2 – 3x +5x – 15) = 1/3 
Or,                    x² + 2x – 15 = 6x + 6 
Or,                            x² – 4x – 21 = 0 
Or,                           (x + 3)(x – 7) = 0

Either, x + 3 = 0   Or, x – 7 = 0

Thus, x = 7 or – 3; but age cannot be negative.

Hence his age is 7 years. 

Example – 15. In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.

Solution :- Let her marks in Mathematics = x, then marks in English = 30 – x. 
As per statements, her marks in Mathematics = x + 2

d her marks in English = 30 – x – 3 = 27 – x 
A/Q                 (x + 2)(27 – x) = 210
Or,                  27x – x² + 54 – 2x = 210 
Or,                  x² – 25x + 156 = 0
Or,                  (x – 12)(x – 13) = 0  
Either,        x – 12 = 0, Or, x – 13 = 0

Thus, x = 12, 13 

Hence, either marks in Mathematics = 12 & marks in English = 18,

Or, marks in Mathematics = 13 & marks in English = 17.

Example – 16. Te diagonal of a rectangular field is 60 m more than the shorter side. If the longer side is 30 m more than the shorter side, find the sides of the field.

Solution :- Lt shorter side be x, then diagonal = x + 60 and longer side = x + 30

By Pythagoras Theorem,    (x + 60)² = (x + 30)² + x². 

Or,        x² + 120x + 3600 = x² + 60x + 900 + x²
Or,       − x² + 60x + 2700 = 0
Or,       x² – 60x – 2700 = 0
Or,       (x – 90)(x + 30) = 0
Either,         x – 90 = 0     Or,     x + 30 = 0

Thus, x = 90 or, x = – 30. But length cannot be negative.

Hence shorter side = 90 m and longer side = 90 + 30 = 120 m.

Example – 17. The difference of squares of two numbers is 180. The square of smaller number is 8 times the larger number. Find the two numbers.

Solution :- Let larger number be x and smaller number y.

The give condition is that,  x² – y² = 180 -----  (1) and y² = 8x -----   (2).

Putting y² = 8x from (2) to (1) we get,  
                                             x² – 8x = 180 
Or,                               x² – 8x – 180 = 0 
Or,                              (x – 18)(x + 10) = 0 
Thus,                                  x = 18  or  – 10 
When x = 18, y² = 8 × 18 = 144, i.e. y = ±12 
Hence, x = 18, y = 12   or   x = 18, y = – 12.

Example – 18. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.

Solution :- Let the original speed of the train = x km/h and increased speed = x + 5 km/h.

The time taken in two cases are : 360/x and 360/(x + 5)

A/Q                                   360/x – 360/(x + 5) = 1
Or,                     [360(x + 5) – 360x] /x(x + 5) = 1
Or,             [360x + 369 x 5 – 360x]/(x² + 5x) = 1
Or,                                            x² + 5x – 1800 = 0

Here, a = 1, b = 5, c = – 1800 ;   and D = b² – 4ac

= 52 – 4 × 1 × (– 1800)  
= 25 + 7200 = 7225

Hence,       x = [– b ± √ D]/ 2a = [– 5 ± √(7225)]/(2 × 1)

= [– 5± 85]/2 = 80/2 or – 90 /2 = 40 or – 45

As speed cannot be negative, therefore, x = 40  or speed = 40 km/h.

Add the following in

Solve the following quadratic equation:

1. x2 - x - 20 = 0 3. 6x2 + 31x + 40 = 0 5. 16x2 - 24x = 0 7.  9. abx2 +(b2 - ac) x - bc = 0

2. 9x2 - 3x - 2 = 0 4. x2 + 6x + 5 = 0 6. 25x2 - 30x + 9 = 0 8.  10. x2 - 4qx + 4q2 = 0

Determine whether the following quadratic equations have real roots and if they have find them:

11.  13.  15.  17.  19.

12.  14. x2 + 3x + 1 = 0 16.  18.  20.

Find the value of k. If the following quadratic equation has equal roots:

21. k²x² - 2(2k - 1) x + 4 = 0

22. x² - 2kx + 7k - 12 = 0

23. ( k + 1) x² - 2(k - 1) x + 1 = 0

24. If the equation (1 + m²) x² + 2 mcx + (c² - a²) = 0 has equal roots, prove that c² = a²(1 + m²)

25. If the roots of the equation (a - b) x² + (b - c) x + (c - a) = 0 are equal, prove that 2a = b + c.

Find the value of p for which the following equations has real roots:

26. 2x² + px + 8 = 0

27. 3x² + 3x + p = 0

28. 5px² - 8x + 2 = 0

Answers
1. -4, 5 4. -1, -5 7. -1 10. 2q 13. -9, 7 16.  19.  22. 3, 4 27.	2. -1/3, 2/3 5. 0, 3/2 8. -2 11. 5, 5/2 14.  17.  20.  23. 0, 3 28.	3. -1/2, 2/3 6. 3/5, 3/5 9. c/b, - b/a 12. 2, -5 15. No real roots. 18.  21. 1/4 26.

Word Problems

1. Student of a class are made to stand in rows. If 4 students are extra in a row, there would be 2 rows less. If 4 students are less in a row, there would be 4 more rows. There would be 4 more rows. Find the number of students in the class.

2. Two pipes running together can fill a cistern in 6 minutes. If one pipe takes 5 minutes more than the other to fill the cistern, find the time in which each pipe would fill the cistern.

3. Rs. 9,000 were divided equally among a certain number of persons. Had there been 20 more persons, each would have got Rs. 160 less. Find the original number of persons.

4. Some student planned a picnic. The budget for food was Rs. 500. But 5 of these failed to go and thus the cost of food for each member increased by Rs 5. How many students attended the picnic?

5. The sum of two number is 15 and sum of there reciprocals is 3/10. Find the numbers.

6. Two numbers differ by 2 and their product is 360. Find the numbers.

7. The length of a rectangle exceeds its width by 8cm and the area of the rectangle is 240sq. cm find the dimensions of the rectangle.

8. The side of a square exceeds the side of another square by 4cm and the sum of the areas of the two squares is 400 sq.cm. find the dimensions of the squares.

9. A train covers a distance of 90 km at a uniform speed. Had the speed been 15km per hour more, it would have taken half an hour less for the journey. Find the original speed of the train.

10. A shopkeeper buys a numbers of books for Rs. 80. If the had bought 4 more books for the same amount, each book would have cost him Rs.1 less. How many books did he buy?

11. A piece of cloth costs Rs 200. If the piece were 5m longer, and each meter of cloth costed Rs. 2 less, the cost of the piece would have remained unchanged. How long is the piece and what is its original rate per meter ?

12. A two digit number is such that the product of its digits is 8. When 63 is subtracted from the number, the digits interchange their places. Find the number.

13. A person on tour has Rs. 360 for his daily expenses. If he extends his tour for 4 days he has to cut down his daily expenses by Rs. 3. Find the original duration of the tour.

14. The sum of the squares of two consecutive natural numbers is 313, find the numbers.

15. Divide 29 into two parts so that the sum of the squares of the parts is 425.

16. Two trains leave a railway stations at the same time. The first train travels due west and the second train due North. The first train travels 5 km/hr faster than the second train. If after 2 hours, they are 50 km apart. Find the average speed of each train.

17. The speed of boat in still water is 8 km/hr. If can go 15km upstream and 22 km downstream in 5 hours. Find the speed of the stream.

18. The sum of the ages, in years, of a son and his father is 35 and the product of their ages is 150. find their ages.

19. The two circles touch internally. The sum of there areas is   sq. cm .and the distance between their centres is 6cm. Find the radil of the circles.

20. Two circles touch externally. The sum of their areas is   . and the distance between their centres is 14cm. Find the radill of the circles.

21. The hypotenuse of a right angled triangle is 6cm more than twice the shortest side. If the third is 2cm less than the hypotenuse, find the sides of the triangle.

22. A year ago, the father was 8 times as old as his son. Now his age is the square of his son’s age. Find their present ages.

23. Find two consecutive even integers, the sum of whose squares is 164.

24. Find two consecutive odd natural numars, the sum of whose squares is 202.

25. The sum of two natural numbers is 8. Determine the numbers if the sum of their reciprocals is 8/15.

26. There are three consecutive positive integers such that the sum of the squares of the first and the product of the other two is 154. What are the integers.

27. A farmer wishes to grow a 100m² rectangular vegetable garden. Since he has with him only 30m barbed wire, he fences three sides of the rectangular garden letting compound wall of his houes act as the fourth side-fence. Find the dimensions of his garden.

28. The sum s of n successive odd natural numbers starting from 3 is given by the relation S = n(n + 2) Determine n if the sum is 168.

29. The sum s of first n even natural number is given by the relation S = n(n + 1) Find n, if the sum is 420.

30. The sum S of first n natural numbers is given by the relation

Find n, if sum is 276.

31. Two water taps together can fill a tank in 9(3/8) hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.  

32. Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares.

33. A pole has to be erected at a point on the boundary of a circular park of diameter 13 m in such a way that the differences of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 m. Is it possible to do so? If yes, at what distances from the two gates should the pole be erected?

Answers
(1) 96 (4) 20 (7) 12cm, 20 cm (10) 16 (13) 20 days (16) 15km/hr, 20km/hr (19) 4cm, 10cm (22) 7 yrs , 49 yrs (25) 3, 5 (28) 12 (31) [15 hours and 25 hours]	(2) 10min, 5min (5) 5, 10 (8) 12cm, 16cm (11) 20m, Rs. 10 per m (14) 12, 13 (17) 3km/hr (20) 11cm, 3cm (23)  (26) 8, 9, 10 (29) 20 (32) [18 m and 12 m]	(3) 25 (6)  (9) 45km/hr (12) 81 (15) 13, 16 (18) 30 yrs, 5 yrs (21) 10cm, 24cm, 26cm (24) 9, 11 (27) 20m X 5m or 10m X 10m (30) 23 (33) [5 m from B and 12 m from A].