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Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations

Previous Year Questions 2024

Q1: In flight of 2800 km, an aircraft was slowed down due to bad weather. Its average speed is reduced by100km/h and by doing so, the time of flight is increased by 30 minutes. Find the original duration of the flight.     (CBSE 2024)

Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations  View Answer

Ans:
Let the speed of aircraft be x km/hr.
Time taken to cover 2800 km by speed of x km/hr = 2800/x  hrs.
New speed is (x – 100) km/hr
so time taken to cover 2800 km at the speed of
Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations
Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations
⇒ 560000 = x2 – 100x
⇒ x2 – 100x – 560000 = 0
⇒ x2 – 800x + 700x – 560000 = 0
⇒ x(x – 800) + 700(x – 800) = 0
⇒ (x – 800) (x + 700) = 0
⇒ x = 800, – 700 (Neglect)
⇒ x = 800
Speed = 800 km/hr
Time = 2800/800
= 3 hr 30 min.


Q2: The denominator of a fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations find the fraction.      (CBSE 2024)

Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations  View Answer

Ans: Let the numerator be x.
Denominator = 2x + 1
Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations
Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations
Let, Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations
Then, the equation will be.
Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations
Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations
⇒ 21y2 + 21 = 58y
⇒ 21y2 – 58y + 21 = 0
⇒ 21y2 – 49y – 9y + 21 = 0
⇒ 7y(3y – 7) – 3(3y – 7) = 0
⇒ (3y – 7) (7y – 3) = 0
Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations
∴ Required fraction will be 7/3 and 3/7.

Previous Year Questions 2023

Q3: Find the sum and product of the roots of the quadratic equation 2x2 - 9x + 4 = 0.  (CBSE 2023)

Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations  View Answer

Ans: Let α and β be the roots of given quadratic equation  2x2 - 9x + 4 = 0.
Sum of roots = α + β = -b/a = (-9)/2 = 9/2
and Product of roots, αβ = c/a = 4/2 = 2  


Q4: Find the value of p, for which one root of the quadratic equation px2 – 14x + 8 = 0 is 6 times the other.  (CBSE 2023)

Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations  View Answer

Ans: Let the first root be α, then the second root will be 6a
Sum of roots  = -b/a
⇒ a + 6a = 14/p
⇒ 7a = 14/p
⇒ a = 2/p
Product of roots = c/a
⇒ a x 6a  = 8/p
⇒ 6a2 = 8/p
Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations
⇒ p = 6 x 4/8
⇒ p = 3
Hence, the value of p is 3.


Q5: The least positive value of k for which the quadratic equation 2x2 + kx + 4 = 0 has rational roots, is (2023)
(a) ±2√2
(b) 4√2
(c) ±2
(d) √2

Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations  View Answer

Ans: (b)
Sol: For rational roots, D is a perfect square
D = k2- 32
If k = 4√2  then,
D = (4√2)2 - 32 = 0. 

Which is a perfect square


Q6: Find the discriminant of the quadratic equation 4x2 - 5 = 0 and hence comment on the nature of roots of the equation.  (CBSE 2023)

Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations  View Answer

Ans: Given quadratic equation is 4x2 - 5 = 0
Discriminant, D = b2 - 4ac = 02 - 4(4)(-5) = 80 > 0
Hence, the roots of the given quadratic equation are real and distinct.


Q7: Find the value of 'p' for which the quadratic equation px(x - 2) + 6 = 0 has two equal real roots.  (2023)

Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations  View Answer

Ans: The given quadratic equation is px(x - 2) + 6 = 0
⇒ px2 - 2xp +6 = 0
On comparing with ax2+ bx + c = 0, we get
a = p, b = -2p and c = 6
Since, the quadratic equations has two equal real roots.

∴ Discriminant D = 0
⇒ b2 - 4ac = 0
⇒ (-2p)- 4 x p x 6 = 0
⇒ 4p- 24p = 0
⇒ p2 - 6p = 0
⇒  p(p - 6) = 0
⇒  p = 0 or p = 6
But p ≠ 0 as it does not satisfy equation
Hence, the value of p is 6.


Q8: Case Study : While designing the school year book, a teacher asked the student that the length and width of a particular photo is increased by n units each to double the area of the photo. The original photo is 18 cm long and 12 cm wide.
Based o n the above information. answer the following Questions:
(i) Write an algebraic equation depicting the above information.
(ii) Write the corresponding quadratic equation in standard form.
(iii) What should be the new dimensions of the enlarged photo?

Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations

OR

Can any rational value of x make the new area equal to 220 cm2 ?      (2023)

Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations  View Answer

Ans: Area = 18 x 12 cm
Length (l) is increased by x cm
So,  new length =(18 + x ) cm
New width = (12 + x) cm
Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations

(i) Area of photo after increasing the length and width
= (18 + x)(12 + x) = 2 x 18 x 12
i.e., (18 + x) (12 + x) = 432 is the required algebric equation.
(ii) From part (i) we get, (18 + x) (12 + x) = 432
 216 + 18x + 12x + x2 = 432
  x2 + 30x - 216 = 0
(iii)  x2 + 30x - 216 = 0
 x2+ 36x - 6x - 216 = 0
  x(x+ 36) - 6 (x+ 36) = 0 ⇒  x = 6, -36
-36 is not possible.
So, new length = (18 + 6) cm = 24 cm
New width = (12 + 6) cm = 18cm
So. new dimension = 24cm x 18 cm

OR

According to question (13 + x) (12 + x) = 220
 216 + 30x + x2 = 220
  x+ 30x + 216 - 220 = 0
  x2 + 30x - 4 = 0
For rational value of x. discriminant (D) must be perfect square.
So, D = b2 - 4ac
= (30)2 - 4(1) (-4) = 900 + 16 = 916
∴  916 is not a perfect square.
So, no rational value of x is possible.


Q9: The roots of the equation x2 + 3x – 10 = 0 are: 
(a)
 2, –5
(b)
 –2, 
(c)
 2, 5
(d)
 –2, –5     (CBSE 2023)

Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations  View Answer

Ans: (a)
To find the roots of the quadratic equation x2 + 3x − 10 = 0, we can use the quadratic formula:

Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations

For the equation x2 + 3x − 10 = 0:
a = 1
b = 3
c = −1
Substitute these values into the formula:

Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations

Now, calculate the two roots:

(i) For x = 3 + 7 / 2 = 4/2 = 2
(ii) For x  = -3 - 7/2 = -10/2 = -5

The roots of the equation are 
2 and -5.
So, the correct answer is: (a) \text{(a) } 2, -5(a) 2,−5

Previous Year Questions 2022

Q10: If the sum of the roots of the quadratic equation ky2 – 11y + (k – 23) = 0 is 13/21 more than the product of the roots, then find the value of k.    (2022)

Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations  View Answer

Ans: Given, quadratic equation is ky2 – 11y + (k – 23) = 0
Let the roots of the above quadratic equation be α and β.
Now, Sum of roots, α + β = -(-11)/k = 11/k ...(i)
and Product of roots, αβ = k-23/k  ...(ii)
According to the question,
α + β = αβ + 13/21
Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations
⇒ 21(34 – k) = 13k
⇒ 714 – 21k = 13k
⇒ 714 = 13k + 21k
⇒ 34k = 714
⇒ k = 714/34
⇒ k = 21


Q11: Solve the following quadratic equation for x: x2 – 2ax – (4b2 – a2) = 0

Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations  View Answer

Ans: x2 – 2ax – (4b2 – a2) = 0 
⇒ x2 + (2b – a)x – (2b + a)x – (4b2 – a2) = 0
⇒ x(x + 2b – a) – (2b + a)(x + 2b – a) = 0
⇒ (x + 2b – a)(x – 2b – a) = 0
⇒ (x + 2b – a) = 0, (x – 2b – a) = 0
∴ x = a − 2b, a + 2b 


Q12: In the picture given below, one can see a rectangular in-ground swimming pool installed by a family In their backyard. There is a concrete sidewalk around the pool of width x m. The outside edges of the sidewalk measure 7 m and 12 m. The area of the pool is 36 sq. m.
Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations

Based on the information given above, form a quadratic equation in terms of x
Find the width of the sidewalk around the pool.
      (2022)

Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations  View Answer

Ans: Given, width of the sidewalk = xm.
Area of the pool = 36 sq.m
∴ Inner length of the pool
= (12 - 2x)m
Inner width of the pool = (7 - 2x)m
∴ Area of the pool. A = l x b
⇒ 36 = (12 - 2x) x  (7 - 2x)
⇒ 36 = 84 - 24x - 14x +4x2
⇒ 4x2 - 38x + 48 = 0
⇒ 2x2 - 19x + 24 = 0, is the required quadratic equation.
Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations
Area of the pool given by quadratic equation is2x2 - 19x + 24 = 0
⇒ 2x2 - 16x - 3x + 24 = 0
⇒  2x(x - 8) - 3(x - 8) = 0
⇒ (x - 8)(2x - 3) = 0
⇒ x = 8 (not possible) or x = 3/2 = 1.5
Width of the sidewalk =1.5m


Q13: The sum of two numbers is 34. If 3 Is subtracted from one number and 2 is added to another. the product of these two numbers becomes 260. Find the numbers.      (2022)

Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations  View Answer

Ans: Let one number be x and another number be y.
Since, x + 4y = 34 ⇒  y = 34 - x      (i)
Now. according to the question. (x - 3) (y + 2) = 260       (ii)
Putting the value or y from (i) in (ii), we get
⇒ (x - 3)(34 - x + 2) = 260
⇒ (x - 3)(36 - x) = 260
⇒ 36x -x2 - 108 + 3x = 260
⇒ x- 39x +368 = 0
⇒ 4x2- 23x - 16x + 368 = 0
⇒ x(x - 23) - 16(x - 23) = 0
⇒ (x - 23)(x - 16) =0
⇒ x = 23 or 16
Hence; when x  = 23 from (i),  y = 3+ - 23 = 11
When x = 16. then y = 34 - 16 = 18
Hence the required numbers are 23 and 11 or 16 and 18.


Q14: The hypotenuse (in cm) of a right angled triangle is 6 cm more than twice the length of the shortest side. If the length of third side is 6 cm less than thrice the length of shortest side, then find the dimensions of the triangle.     (2022)

Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations  View Answer

Ans: 

Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations
Let ΔABC be the right angle triangle, right angled at B, as shown in the figure.
Also, let AB = c cm, BC = a cm and AC = b cm
Then, according to the given information, we have
b = 6 + 2a  .....(i) (Let a be the shortest side)
and c = 3a – 6  ...(ii)
We know that, b2 = c2 + a2
⇒ (6 + 2a)2 = (3a – 6)2 + a2 ...[Using (i) and (ii)]
⇒ 36 + 4a2 + 24a = 9a2 + 36 – 36a + a2
⇒ 60a = 6a2
⇒ 6a = 60  ...[∵ a cannot be zero]
⇒ a = 10 cm
Now, from equation (i),
b = 6 + 2 × 10 = 26
and from equation (ii),
c = 3 × 10 – 6 = 24
Thus, the dimensions of the triangle are 10 cm, 24 cm and 26 cm.


Q15: Solve the quadratic equation: x2 – 2ax + (a– b2) = 0 for x.     (2022)

Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations  View Answer

Ans: We have, x2 – 2ax + (a2 – b2) = 0
⇒ x2 – ((a + b) + (a – b))x + (a2 – b2) = 0
⇒ x2 – (a + b)x – (a – b)x + (a + b)(a – b) = 0  ......[∵ a2 – b2 = (a + b)(a – b)]
⇒ x(x – (a + b)) – (a – b)(x – (a + b)) = 0
⇒ (x – (a + b))(x – (a – b) = 0
⇒ x = a + b, a – b 


Q16: Find the value of m for which the quadratic equation (m - 1) x2 + 2 (m - 1) x + 1 = 0 has two real and equal roots.   (2022)

Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations  View Answer

Ans:  We have
(m - 1) x2 + 2 (m - 1) x + 1 = 0 ----(i)
On comparing the given equation with ax2 + bx + c = 0,
we have a = (m - 1), b = 2 (m - 1), c = 1
Discriminant, D = 0
⇒ b2 - 4ac = 0 ⇒ 4m2 + 4 - 8m - 4m + 4 = 0
⇒ 4m2 -12m + 8 = 0
⇒ m2 - 3m + 2 = 0 ⇒ m2 - 2m - m + 2= 0
⇒ m(m - 2) - 1 (m - 2) = 0
⇒ (m - 1)(m - 2) = 0 ⇒ m = 1, 2


Q17: The quadratic equation (1 + a2)x2 + 2abx + (b2 - c2) = 0 has only one root. What is the value of c2(1 + a2)?    (2022)

Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations  View Answer

Ans: (1 + a2)x2 + 2abx + (b2 - c2)= 0
Comparing on Ax2 + Bx + C = 0
A = 1 + a2, B = 2ab & C = (b2 - c2)
Now, B2 - 4AC = 0
⇒ (2ab)2 - 4 × (1 + a2) × (b2 - c2) = 0
⇒ 4a2b2 - 4(b2 - c2 + a2b- a2c2) = 0
⇒ 4a2b2 - 4b2 + 4c2 - 4a2b2 + 4 a2c2 = 0
⇒ - b2+ c2 + a2c2 = 0
⇒ c2 + a2c2 = b2
∴ c2 (1 + a2) = b2

Previous Year Questions 2021


Q18: Write the quadratic equation in x whose roots are 2 and-5.     (2021)

Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations  View Answer

Ans: Roots of quadratic equation are given as 2 and - 5.
Sum of roots = 2 + (-5) = -3
Product of roots = 2 (-5) = -10
Quadratic equation can he written as
x2 - (sum of roots)x + Product of roots = 0
⇒ x2 + 3x - 10 = 0


Previous Year Questions 2020


Q19: Sum of the areas of two squares is 544 m2. If the difference of their perimeters is 32 m, find the sides of the two squares.     (2020)

Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations  View Answer

Ans: Let the sides of the two squares be x m and y m, where ; x > y.
Then, their areas are xand y2 and their perimeters are - 4x and 4y respectively.
By the given condition, x2 + y= 544 --------(i)
and 4x - 4y = 32
⇒ x - y = 8
⇒ x = y + 8 ------------ (ii)
Substituting the value of x from (ii) in (i) we get
⇒ (y + 8)2 + y2 = 544
⇒ y2 + 64 + 16y + y= 544
⇒ 2y2 + 16y - 480 = 0
⇒ y2 + 8y - 240 = 0
⇒ y2 + 20y - 12y - 240 = 0
⇒ y(y + 20) - 12(y + 20) = 0
⇒ (y - 12) (y + 20) = 0
⇒ y = 12    (∵ y ≠ 20 as length cannot be negative)
From (ii), x = 12 + 8 = 20 Thus, the sides of the two squares are 20 m and 12 m.


Q20: A motorboat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.     (2020)

Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations  View Answer

Ans: Let the speed of the stream be x km/hr.
Speed of the boat upstream = (18 - x) km/hr
Speed of the boat downstream = (18 + x) km/hr
According to question,

(∵ x ≠ -54 as speed can’t be negative)
Hence, the speed of the stream is 6 km/hr.


Q21: The value(s) of k for which the quadratic equation 2x2 + kx + 2 = 0 has equal roots, is
(a) 4
(b) ± 4
(c) - 4
(d) 0   (2020, 1 Mark)

Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations  View Answer

Ans: (b)
Given Quadratic equation is 2x2 + kx + 2 = 0
Since, the equation has equal roots.
∴ Discriminant = 0
⇒ k- 4 x 2 x 2 = 0
⇒ k2- 16 = 0
⇒ k2 = 16
⇒ k = ±4


Q22: Solve for Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations (CBSE 2020)

Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations  View Answer

Ans: Given, Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations
Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations
⇒ (x + 4)(x – 7) = –30 
⇒ (x + 4) (x – 7) + 30 = 0 
⇒ x2 + 4x – 7x – 28 + 30 = 0 
⇒ x2 – 3x – 28 + 30 = 0 
⇒ x2 – 3x + 2 = 0 
⇒ x2 – 2x – x + 2 = 0 
⇒ x(x – 2) – 1(x – 2) = 0 
⇒ (x – 2) (x – 1) = 0 
i.e., x – 1 = 0 or x – 2 = 0 
⇒ x = 1 or 2


Q23: A train covers a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, it would have taken 3 hours more to cover the same distance. Find the original speed of the train. (CBSE 2020)

Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations  View Answer

Ans: Let the original speed of the train be x km/h. Then, time taken to cover the journey of 480 km = 480 / x hours 
Time taken to cover the journey of 480 km with speed of (x – 8) km/h = 480 / x − 8 hours 
Now, according to question,
Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations

Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations
⇒ 3x(x – 8) = 3840 
⇒ x(x – 8) = 1280 
⇒ x2 – 8x – 1280 = 0 
⇒ x2 – 40x + 32x – 1280 = 0 
⇒ x(x – 40) + 32(x – 40) = 0 
⇒ (x + 32) (x – 40) = 0 
⇒ x + 32 = 0 or x – 40 = 0 Q 
∴ x = – 32 (not possible) 
∴ x = 40 Thus, the original speed of the train is 40 km/h.

Previous Year Questions 2019

Q24: Find the value of k for which x = 2 is a solution of the equation kx2 + 2x - 3 = 0.     (CBSE 2019)

Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations  View Answer

Ans: Since x = 2 is a solution of kx2 + 2x - 3 = 0
k(2)2 + 2(2) - 3 = 0
= 4k + 4 - 3 = 0
⇒ k = -1/4


Q25: Sum of the areas of two squares is 157 m2. If the sum of their perimeters is 68 m, find the sides of the two squares.    (CBSE 2020)

Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations  View Answer

Ans: Let the length of the side of one square be x m and the length of the side of another square be y m.
Given, x2 + y2= 157    _(i)
and 4x + 4y=68    _(ii)
 x + y = 17
y = 17 - x    _(iii)
On putting the value of y in (i), we get
x+ (17 - x)2 = 157
⇒ x2 + 289 + x2 - 34x = 157
=> 2x2 - 34x +132 = 0
⇒ x2 - 17x + 66 = 0
⇒ x2 - 11x - 6x + 66 =0
⇒ x(x - 11)-6(x - 11) = 0
⇒ (x - 11) (x - 6) = 0
⇒ x = 6 or x = 11
On putting the value of x in (iii), we get
y = 17 - 6 = 11 or y = 17 - 11 = 6
Hence, the sides of the squares be 11 m and 6 m.

Previous Year Questions 2017

Q26: If one root of the quadratic equation 6x2 – x – k = 0 is 2/3, then find the value of ‘k’.  (CBSE 2017)

Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations  View Answer

Ans: Given quadratic equation is: 
6x2 – x – k = 0. 
Its one root is 2/3
If x = 2 / 3 is root of the given equation, then it will satisfy the given equation,
Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations
⇒ k = 2
Hence, the value of k is 2.


Q27: If the equation (1 + m2)x2 + 2 mcx + c2 – a2 = 0 has equal roots then show that c= a2 (1 + m2).  (CBSE 2017)

Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations  View Answer

Ans: Given, quadratic equation is, 
(1 + m2 )x2 + 2mc x + c2 – a2 = 0 
On comparing it with Ax2 + Bx + C = 0, 
we get A = 1 + m2, B = 2mc and C = c2 – a2
The roots of the given equation are equal, then 
Discriminant, D = 0 
∴ B2 – 4AC = 0 
(2 mc)2 – 4 × (1 + m2) × (c2 – a2) = 0 
⇒ 4m2c2 – 4(c+ c2m– a2 – a2m2) = 0 
⇒ 4m2c2 – 4c– 4c2m2 + 4a2 + 4m2a2 = 0 
⇒ m2a2 + a2 – c2 = 0 
⇒ c2 = m2a2 + a2
⇒ c2 = a2 (1 + m2
Hence, proved


Q28: A line segment AB of length 2 m is divided at a point C into two parts such that AC2 = AB × CB. Find the length of CB. (CBSE 2017)

Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations  View Answer

Ans: Let the length of AC be x 
Then, BC = (2 – x) m 
Now, AC2 = AB × CB (given) 
∴ x2 = 2 × (2 – x) [ AB = 2 m (given)] 
⇒ x2 = 4 – 2x 
⇒ x+ 2x – 4 = 0 
Now, if we compare the above equation with ax2 + bx + c = 0. 
Then, a = 1, b = 2, c = –4 
Roots of the equation are
Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations
Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations
Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations (which is not possible) 
Hence, the length of BC is 3 − √5 or 0.76 m.

Previous Year Questions 2011

Q29: The roots of the equation x2 – 3x – m(m + 3) = 0, where m is a constant, are:    
(a) m, m  + 3 
(b) –m, m + 3
(c)
 m, –(m + 3) 
(d) –m, –(m + 3) (CBSE 2011)

Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations  View Answer

Ans: (b)
To find the roots of the equation x^2 - 3x - m(m + 3) = 0x2 − 3x − m(m + 3) = 0, we can use the quadratic formula:

Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations

For the equation x^2 - 3x - m(m + 3) = 0x2 − 3x − m(m + 3) = 0:
a = 1
b = −3
c = −m(m + 3)
Substitute these values into the formula:

Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations

Since Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations , we get two cases for the roots based on the sign:
Case 1: Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations
Case 2: Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations
Thus, the roots of the equation are -m and m + 3.
So, the correct answer is: (b) −m, m + 3

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FAQs on Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations

1. What are quadratic equations and how do they differ from linear equations?
Ans. Quadratic equations are polynomial equations of the form ax² + bx + c = 0, where a, b, and c are constants, and a ≠ 0. They differ from linear equations, which are of the form y = mx + b, where m and b are constants. Linear equations graph as straight lines, while quadratic equations graph as parabolas.
2. How can I solve a quadratic equation?
Ans. A quadratic equation can be solved using various methods, including factoring, completing the square, or applying the quadratic formula: x = (-b ± √(b² - 4ac)) / (2a). The choice of method depends on the specific equation and its coefficients.
3. What are the types of roots of a quadratic equation?
Ans. The roots of a quadratic equation can be real and distinct, real and equal, or complex. The nature of the roots is determined by the discriminant (b² - 4ac): if it's positive, there are two distinct real roots; if it's zero, there is one real root; if it's negative, the roots are complex.
4. How do quadratic equations apply in real-life scenarios?
Ans. Quadratic equations are used in various real-life situations, such as calculating areas, projectile motion in physics, and optimizing profits in business. They help model situations where relationships between variables are not linear.
5. What is the significance of the vertex in a quadratic equation?
Ans. The vertex of a quadratic equation represents the maximum or minimum point of the parabola, depending on whether it opens upwards or downwards. It is significant in understanding the behavior of the quadratic function and helps in graphing the equation.
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