CBSE Class 10 Mathematics Quadratic Equation Worksheet Set L

Read and download free pdf of CBSE Class 10 Mathematics Quadratic Equation Worksheet Set L. Students and teachers of Class 10 Mathematics can get free printable Worksheets for Class 10 Mathematics Chapter 4 Quadratic Equation in PDF format prepared as per the latest syllabus and examination pattern in your schools. Class 10 students should practice questions and answers given here for Mathematics in Class 10 which will help them to improve your knowledge of all important chapters and its topics. Students should also download free pdf of Class 10 Mathematics Worksheets prepared by school teachers as per the latest NCERT, CBSE, KVS books and syllabus issued this academic year and solve important problems with solutions on daily basis to get more score in school exams and tests

Worksheet for Class 10 Mathematics Chapter 4 Quadratic Equation

Class 10 Mathematics students should refer to the following printable worksheet in Pdf for Chapter 4 Quadratic Equation in Class 10. This test paper with questions and answers for Class 10 will be very useful for exams and help you to score good marks

Class 10 Mathematics Worksheet for Chapter 4 Quadratic Equation

 

Quadratic Equation

Q.- Solve :

(i) x +5/x
(ii) 3y + 5/16y = 2
 
Sol.
(i) x +5/x = 6
=> x2 + 5 = 6x [Multiplying each term by x]
=> x2 – 6x + 5 = 0
=> x2 – 5x – x + 5 = 0
i.e., x(x – 5) – 1(x – 5) = 0
=> (x – 5) (x – 1) = 0 i.e., x – 5 = 0
or x – 1 = 0
=> x = 5 or x = 1.
∴ Required solution is 5, 1
 
(ii) 3y +5/16y = 2
=> 3y × 16y + 5 = 2 × 16y
=> 48y2 – 32y + 5 = 0
=> 48y2 – 12y – 20y + 5 = 0
i.e., 12y (4y – 1) – 5(4y – 1) = 0
=> (4y– 1) (12y – 5) = 0
i.e., 4y – 1 = 0 or 12y – 5 = 0
=> 4y = 1 or 12y = 5 i.e., y = 1/4 or y = 5/12
∴ Required solutions is : 1/4,5/12
 
Q.- Solve :
(i) √x + 2x = 1
(ii) 3x2 - 2 + 1 = 2x
(iii) 2x2 + 9 + x = 13
 
Sol.
(i) √x + 2x = 1
=> √x = 1 – 2x
i.e., x = (1 – 2x)2
=> x = 1 + 4x2 – 4x
i.e., 1 + 4x2 – 4x – x = 0
=> 4x2 – 5x + 1 = 0 i.e., 4x2 – 4x – x + 1 = 0
=> 4x (x – 1) – 1 (x – 1) = 0
i.e., (x – 1) (4x – 1) = 0
=>x – 1 = 0
or 4x – 1 = 0
i.e., x = 1 or x = 1/4
 
Q.- Find two consecutive natural numbers, whose product is equal to 20.
 
Sol. Let the required two consecutive natural numbers be x and x + 1.
Given : x (x + 1) = 20
=> x2 + x – 20 = 0
=> (x + 5) (x – 4) = 0
=>  x = – 5, or x = 4
Since, x must be a natural number,
∴ x = 4
And required numbers are x and x + 1 i.e., and 5.
 
Q.- The sum of the squares of two consecutive whole numbers is 61. Find the numbers.
 
Sol. Let the required consecutive whole numbers be x and x + 1.

∴  x2 + (x + 1)2 = 61

=> x2 + x2 + 2x + 1 – 61 = 0
=> 2x2 + 2x – 60 = 0
=>  x2 + x – 30 = 0 [Dividing each term by 2]
=>  (x + 6) (x – 5) = 0 [On factorising]
=> x = – 6, or x = 5
x is a whole number, ∴ x = 5
And, required numbers are x and x + 1 = 5 and 5 + 1 i.e., 5 and 6
 
Q.- Divide 16 into two parts such that twice the square of the larger part exceeds the square of the smaller part by 164.
 
Sol. Let larger part be x, therefore the smaller part
= 16 – x
Given : 2x2 – (16 – x)2 = 164
=> 2x2 – (256 + x2 – 32x) – 164 = 0
i.e., 2x2 – 256 – x2 + 32x – 164 = 0
=> x2 + 32x – 420 = 0
On factorizing, it gives : (x + 42) (x – 10) = 0
i.e., x = – 42 or x = 10
∴ x = 10
Hence the larger part = 10 and the smaller part
= 16 – x = 16 – 10 = 6
 
Q.- Two positive numbers are in the ratio 2 : 5. If difference between the squares of these numbers is 189 ; find the numbers.
 
Sol. Let numbers be 2x and 5x
∴ (5x)2 – (2x)2 = 189
=> 25x2 – 4x2 = 189 and 21x2 = 189
i.e., x2 = 189/21 = 9
=> x = ± 3
Since, the required numbers are positive,
∴ x = 3
And, required numbers = 2x and 5x = 2 × 3 and
5 × 3 = 6 and 15

 

More question- 

1) Find the discriminate of the quadratic equation: 3 √3 x2 + 10x + √3 = 0 (64)

2) Solve for x: a) 9x2 – 9 (a + b) x + 2a2 + 5ab + 2b2 = 0 (2a+b/3, a + 2b/3)

b) 4x2 – 4a2x + (a4– b4) = 0 (a2 +b2)/2, (a2- b2)/2

c) 10ax2 – 6x +15ax – 9 = 0 (-3/2, 3/5a)

d) x2 – 2(a2 + b2 )x + (a2 – b2)2 = 0 (a+b)2, (a-b)2

e) √7x2 – 6x – 13 √7 (13√7/7, - √7)

f) x2 - 5√5x + 30 = 0 (3√5, 2√5)

3) find the value of k so that the quadratic equation has equal roots:

a) 2kx2 – 40x + 25 = 0 (k = 8)

b) 2x2 – (k – 2) x + 1 = 0 (5, 7)

c) ( k + 3 ) x2 + 2 ( k + 3 )x + 4 = 0

4) For what value of p the equation (1 + p) x2 + 2(1 + 2p) x + (1 + p) = 0 has coincident roots (0, -2/3)

5) Find the roots of the following quadratic equation by the method of completing the Square.

a) a2 x2 – 3abx + 2b2 = 0

b) x2 – 4ax + 4a2- b2= 0 c) 6x2 – 7x + 2 = 0 (2/3, ½)

6) Solve the following quadratic equations by factorization method:

a) 3x2 - 2√6x + 2 = 0 (√2/3, √2/3)

7) write the nature of roots of quadratic equation: 4x2 + 4√3x + 3 = 0

8) Check whether the equation x3 – 4x2 + 1 = (x – 2)2 is quadratic or not

9) Solve for x: 1 = 1/a+b+x = 1/a+ 1/b+ 1/x          a + b ≠ 0  (-a,-b)

10) If p, q are the roots of the equation x2 – 5x + 4 =0, find the value of 1/p + 1/q - 2pq  (-27/4)

11) Solve for x: x / x +1  +x +1/x = 34/15     (3/2, -5/2)

12) Solve for x: 1/x – 3 - 1/x+5  = 1/6  (7,-9)

13) If one root of a quadratic equation 3x2+ PX + 4 = 0 is 2/3, find the value of p (p = -8)

14) If x = √2 is a solution of quadratic equation x2 + k x – 4 = 0, then find the value of k

15) Solve for x: 

CBSE Class 10 Quadratic Equation (5) 1

 (-10, -1/5)

16) Solve the equation: 2(x – 3)2 + 3(x – 2) (2x – 3) = 8(x + 4) (x – 4) - 1 (x =5)

17) If the roots of the equation (b – c) x2 + (c – a) x + (a – b) = 0 are equal, then prove that 2b = a + c

18) The sum of the squares of two consecutive odd numbers is 394. Find the numbers. (13, 15)

19) Find two consecutive numbers, whose squares have the sum 85. (6, 7)

20) The product of 3 consecutive even numbers is equal to 20 times their sum. Find the numbers (6, 8, and 10)

21) The sum of the areas of two squares is 640 m2. If the difference in their perimeter is 64m .Find the sides of the two squares (8m, 24m)

22) The difference of two numbers is 4. If the difference of their reciprocals is 4/21, find the numbers (3, 7)

23) The sum of two numbers is 15 and sum of their reciprocals is 3/10. Find the numbers (5, 10)

24) A plane left 30 minutes late than its scheduled time and in order to reach the destination 1500km away in time it had to Increase the speed by 250 km/h from the usual speed. Find its usual speed (750 km / hr)

25) The hypotenuse of a grassy land in the shape of a right triangle is 1m more than twice the shortest side. If the third side is 7m More than the shortest side find the sides of grassy land ( 8, 15)

26) The perimeter of a right angled triangle is 70units and its hypotenuse is 29 units. Find the lengths of the other sides (20, 21)

27) The length of the sides forming a right angled Δ is 5x cm and (3x – 4) cm. Area of the triangle is 60 cm2. Find the hypotenuse (17cm)

28) The length of the hypotenuse of a right angled triangle exceeds the base by 1cm and also exceeds twice the length of the altitude by 3cm. Find the length of each side of Δ (base = 12cm, hyp = 13cm, altitude = 5cm)

29) A natural number, when increased by 12, becomes equal to 160 times its reciprocal. Find the number

 

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