Most Important Questions Class 10 Maths Quadratic Equation with Solutions

We all know the importance of math, be it in school or real life. We acknowledge that math is a fundamental skill that aids in a variety of tasks. Quadratic Equation Important Questions include the must-practice questions for students for the best in their classes.

Mathematics can be a challenging subject for many students, especially when it comes to complex topics like quadratic equations. However, with the right preparation and practice, you can master this essential topic in Class 10 math. Here, we will guide you through the most important quadratic equation questions, explaining their significance, how to approach them, and how extra practice can benefit your studies.

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Chapter 4  Quadratic Equations: Important Questions

1. Aman solved a quadratic equation and found its roots to be real.

Which of these could represent the graph of the equation Aman solved?

a. only iii)

b. only i) and ii)

c. only iii) and iv)

d. only i), ii) and iv)

Soln. 

(d) only i), ii) and iv)

Explanation: 

We must analyse each graph to determine which graph represents the quadratic equation with real roots. A quadratic equation has real roots if its parabola intersects the x-axis.

Graph (i):

• The parabola opens downwards.
• The graph intersects the x-axis at two distinct points.
• This means the equation has two distinct real roots.
• Graph (i) could represent the equation Aman solved.

Graph (ii):

• The parabola opens downwards.
• The graph touches the x-axis at one point (vertex).
• This means the equation has one real root (a repeated root).
• Graph (ii) could represent the equation Aman solved.

Graph (iii):

• The parabola opens upwards.
• The graph does not intersect the x-axis.
• This means the equation has no real roots.
• Graph (iii) cannot represent the equation Aman solved.

Graph (iv):

• The parabola opens upwards.
• The graph touches the x-axis at one point (vertex).
• This means the equation has one real root (a repeated root).
• Graph (iv) could represent the equation Aman solved.

The graphs that could represent the quadratic equation Aman solved are:

• Graph (i) (two distinct real roots),
• Graph (ii) (one real root),
• Graph (iv) (one real root).

2. Three students were asked how they would verify their solution of a quadratic equation, (x-2)(x-5)= 0. Shown below are their responses.

Student 1 said, "In the first bracket, x must equal 2, and in the second bracket, equal 5. So, (2-2)(5 - 5) = 0."

Student 2 said, "In the first bracket, x must equal 2, but in the second bracket, have any real number value. For example, (2-2)(3-5)= 0 or (2-2)(10-5) = 0

Student 3 said, "Both brackets should always have the same x value. So, x is ei or 5 in both brackets. For example, (2-2)(2 - 5) = 0 and (5-2)(5-5) = 0.

"Whose response is correct?

a. only student 1

b. only student 3

c. only students 1 and 2

d. all students 1, 2 and 3

Soln. 

(a) only student 1

Explanation: 

Let’s carefully analyze the responses of all three students with respect to the given quadratic equation:

(x−2)(x−5)=0

For (x−2)(x−5)=0, the product will be zero when:

1. x−2=0  ⟹  x=2, or
2. x−5=0  ⟹  x=5

Student 1:

Student 1 says, "In the first bracket, x must equal 2, and in the second bracket, x must equal 5.

So, (2−2)(5−5)=0"

• Correct interpretation: Student 1 correctly identifies the two solutions, x=2 and x=5, and substitutes them into the equation to verify.
• Verdict: Student 1 is correct

Student 2:

Student 2 says, "In the first bracket, x must equal 2, but x can have any real value in the second bracket. For example, (2−2)(3−5)=0 or (2−2)(10−5)=0"

• Incorrect interpretation: The zero product property states that one of the brackets must equal zero, but the other factor does not need to be zero. Student 2 misinterprets this as x being allowed to have different values in the second bracket, which is invalid for solving the equation.
• Verdict: Student 2 is incorrect.

Student 3:

Student 3 says, "Both brackets should always have the same x value. So, x is either 2 or 5 in both brackets. For example,(2−2)(2−5)=0 and (5−2)(5−5)=0"

• Incorrect interpretation: The two brackets represent different conditions for x to satisfy the equation. It is not required for both brackets to have the same x value simultaneously.
• For x=2, the first bracket is zero. For x=5, the second bracket is zero.
• The statement x must be the same for both brackets is invalid.
• Verdict: Student 3 is incorrect.

3. Arpit was asked to represent the following statements in the form of a quadratic equation:

"The sum of the squares of two positive integers is 225. The square of the larger number is 16 times the smaller number."

If Arpit wrote the equation correctly, what could he have written?

Soln.  

x = the smaller number (a positive integer),

y = the larger number (a positive integer).

"The sum of the squares of two positive integers is 225":
Mathematically:

x²+y²=225

"The square of the larger number is 16 times the smaller number":
Mathematically:

y²=16x

Substitute y²=16x in x²+y²=225

x²+16x=225.

x²+16x−225=0.

This is the required quadratic equation.

4. Find the solution(s) of the following equation.

Show your steps and give valid reasons.

Soln.  

Simplify ( (1/y-3)+(2y/y-1))

(y−1)+(2y²−6y)​/(y−3)(y−1)

(2y²−5y-1)​/(y−3)(y−1)

Now original equation

(y-1) (y-3) (y−1)+(2y²−5y-1)​/(y−3)(y−1) =  2

(since y≠1,3)

2y²−5y−1=2

2y²−5y−1−2=0

2y²−5y−3=0

2y²−3y-2y−3=0

y= -1/2 or y = 3

(since y≠1,3)

y= -1/2

5. One of the solutions of the following equation is -7 where k is a constant.

z² - kz - 28 = 0

i) Find the value of k.

ii) Find the other solution

Show your steps.

Soln.  

i) Find the value of k.

z²−kz−28=0

z=−7

(−7)²−k(−7)−28=0

49+7k−28=0

49−28+7k=0

21+7k=0

7k=−21

k=−3

ii) Find the other solution.

The equation is now:

z²−(−3)z−28=0 or  z²+3z−28=0.

z= −b±√b²−4ac​ /2a

z= −(3)±√(3)²−4(1)(-28)​ /2(1)

z= −3±√121​ /2

z= −3±11​ /2

z= 4 or z = -7

6. A quilt maker has a rectangular quilt measuring 12 units by 8 units. He wants to add a border to it as shown in the figure below. He has 64 sq units of fabric for the border.

(Note: The figure is not to scale.)

i) If x and 2 x are the widths of the border as shown, frame a quadratic equation

using the total area enclosed by the new quilt (with the border).

ii) Find the measures of the new quilt (with the border).

Show your work along with valid reasons.

Soln.  

i) If x and 2 x are the widths of the border as shown, frame a quadratic equation

using the total area enclosed by the new quilt (with the border).

Width=8+x+x=8+2x

Height=12+2x+2x=12+4x.

Total Area=Width×Height.

Total Area=(8+2x)(12+4x)

Area of Original Quilt=12×8=96square units.

Area of Border=Total Area−Area of Original Quilt

Total Area−96=64

(8+2x)(12+4x)−96=64

(8+2x)(12+4x)=8(12)+8(4x)+2x(12)+2x(4x) (distributive property)

(8+2x)(12+4x)=96+32x+24x+8x².

(8+2x)(12+4x)= 8x²+56x+96

Now substitute this into the equation:

8x²+56x+96 = 64

8x²+56x= 64

8x²+56x−64=0

x²+7x−8=0

x= −b±√b²−4ac / 2a

x= −(7)±√(7)²−4(1)(-8) / 2(1)

x= −7±9​ / 2

x= 1 or x = -8

ii) Find the measures of the new quilt (with the border).

New width:

8+2x=8+2(1)=10units.

New height:

12+4x=12+4(1)=16 units.

7. Study the given information and answer the questions that follow.

Bangalore city corporation is building parks in residential areas across the city. Shown below is one such park. The rectangular park consists of various components like walking track, kids play area, open gym, pond etc.

(Note: The image is not to scale.)

(i) Gate 3 has been placed exactly opposite to gate 1 on the boundary of the park. The distance between gate 3 and gate 2 is 1 m more than the distance between gate 3 and gate 2.

The shortest distance between gates 1 and 2 is 29 m, find the width of the park. Show your work.

(ii) The caretaker of the park is attempting to plant saplings in the form of a square. That is, number of rows of saplings is the same as the number of columns of saplings. On arranging the saplings, he found that 24 saplings were still left with him. When he increased the number of rows and columns by 1, he found that he was short of 25 saplings.

Find the number of saplings available with him. Show your work.

Soln.  

(i) Gate 3 has been placed exactly opposite to gate 1 on the boundary of the park. The distance between gate 3 and gate 2 is 1 m more than the distance between gate 3 and gate 2.

The shortest distance between gates 1 and 2 is 29 m, find the width of the park. Show your work.

Let the width of the park be denoted as W. The length of the park is given as 26 + 1 = 27 m. The configuration tells us that Gate 1 and Gate 3 are opposite each other along the longer side (length), while Gate 2 is positioned such that it maintains the shortest distance of 29 m to Gate 1.

From the problem, we know:

• The distance between Gate 3 and Gate 1 is equal to W (since they are on the same side).
• The distance between Gate 3 and Gate 2 is W + 1 m.

Using the information about distances, we can set up the equation for the shortest distance between Gate 1 and Gate 2:

Distance(Gate 1 to Gate 2) = Distance(Gate 3 to Gate 1) + Distance(Gate 3 to Gate 2)
29 m = W + (W + 1)

Simplifying the equation:

29 = W + W + 1

29 = 2W + 1

28 = 2W

W = 14

The width of the park is 14 meters.

(ii) The caretaker of the park is attempting to plant saplings in the form of a square. That is, number of rows of saplings is the same as the number of columns of saplings. On arranging the saplings, he found that 24 saplings were still left with him. When he increased the number of rows and columns by 1, he found that he was short of 25 saplings.

Find the number of saplings available with him. Show your work.

Let N be the number of rows (and columns) of saplings that the caretaker originally planned to plant. The total number of saplings he needs is N². According to the problem, he has 24 saplings left after planting:

N² + 24 = Total Saplings Available

Next, when he increases the rows and columns by 1:
(N + 1)² = Total Saplings Available - 25

From the new arrangement:

(N + 1)² = N^2 + 25

Expanding (N + 1)^2 gives us:
N² + 2N + 1 = N² + 25

Now, simplifying:
2N + 1 = 25
2N = 24
N = 12

The caretaker initially planned to plant 12 rows and 12 columns of saplings.

8. 3(3)2^m + 11(3)^m = 4

Use the substitution 3^m = x to solve for m. Show your steps.

Soln.  

3^2m=(3^m)²=x².

3x²+11x=4.

3x²+11x−4=0

x= (−b±√b²−4ac)/2a

x= (−11±√169)/2(3)

x= (−11±13)/6

x= ⅓ or x=-4

Recall that x=3^m. Substituting back:

⅓ =3^m

3^-1 =3^m

m = -1

For x=−4:

3^m=−4

However, 3^m (an exponential function with base 3) can never be negative because the exponential function is always positive.

Therefore, this solution is not valid.

9. (2 x + 1)³ = 8x (x² + 1) + 3

Write the zeroes of the above equation in the form (x,y). Show your steps.

Soln.  

(2x+1)³−8x(x²+1)−3=0

(2x+1)³=(2x)³+3(2x)²(1)+3(2x)(1²)+1³ (Binomial Theorem)

(2x+1)³=8x³+12x²+6x+1.

(8x³+12x²+6x+1)−(8x³+8x)−3=0

After simplifying

12x²−2x−2=0.

x= (2±10)/4

x= ½ or x= -⅓

Since the corresponding y-value is (x, y)(x,y) are:

( ½, 0) or ( -⅓,0)

10. The graph below represents the path of a ball thrown by Ankush. The maximum height, h, the ball reaches with respect to time, t, is represented by the polynomial h (t) = t²+ 19/4t + 5/4

(Note: The figure is not to scale.)

How long will it take for the ball to reach the ground? Show your steps.

Soln.  

h (t) = t²+ 19/4t + 5/4

t²+ 19/4t + 5/4 = 0

4t²+19t+5=0.

t= −b±√b²−4ac​ /2a

where a=4 b=19, and c=5.

Substituting the values

t= −(19)±√(19)²−4(4) (5)​ /2 (4)

t= −(19)±√361-80​ /8

t= −(19)±√281 /8

t= −(19)±16.76 /8

t≈−0.28 or ≈−4.47 seconds

t≈4.47 seconds.

It will take approximately 4.47 seconds for the ball to reach the ground.

What is the Quadratic Equation Class 10?

A quadratic equation is an equation of the second degree, which means it includes a variable raised to the power of two. It typically takes the form:

ax²+bx+c=0

In this equation:

• a, b, and c are constants (with a≠0);
• The variable is x.

Quadratic equations are a fundamental part of algebra and appear frequently in both academic studies and real-world applications.

Importance: Class 10 Important Questions

Practice is never completed in subjects like math. Class 10 math: important questions are the best to help students make their concepts strong. Here’s why extra questions are essential:

• Extra questions range from basic to complex, enabling you to delve into various facets of quadratic equations. This deepens your understanding and solidifies your knowledge of the topic.
• By practicing a variety of questions, you become better at finding patterns and solving problems more efficiently.
• Solving extra questions helps you gain confidence as you get used to different types of problems that may appear in exams.
• The concepts learned in Class 10 serve as the basis for higher-level mathematics.  Extra questions help ensure you're well-prepared for future challenges.
• The more you practice, the better you get. Extra questions can expose you to different problem-solving methods, helping you perform better in exams.

Benefits of Solving Important Quadratic Equation Questions

Practicing important questions offers multiple benefits beyond just helping you score well on exams. Here’s how these questions can be beneficial:

• Working through important questions revises the lessons you've learned in class, helping to reinforce those concepts in your mind.
• Many important questions require you to think critically and analytically, skills that are not only important for math but for problem-solving in general.
• Regular practice with these questions helps improve your speed and accuracy, both of which are essential during timed exams.
• By practicing a variety of questions using Quadratic Equation Class 10 Important Questions with Solutions, you start to develop strategies for approaching different types of problems, which can be incredibly useful in an exam setting.
• Identifying your weaknesses during practice lets you focus on them and improve before the exam.

How to Solve 10 Important Class Questions Effectively?

When preparing for your exams, it's essential to know how to approach important questions effectively. Here are some steps you can follow:

• Start by ensuring you fully understand the concept behind quadratic equations. Know how to factor, use the quadratic formula, and complete the square.
• Different types of questions require different approaches. Determine the question and solve using the quadratic formula or another method.
• As you read the question, note down the key information provided, such as the values of a, b, and c, or any specific conditions mentioned.
• Based on the type of question, choose the most suitable method to solve it. For example, if the equation can be easily factored, then factorization is the way to go.
• Break down the problem into manageable steps, solving each one carefully. Avoid skipping steps, as this can lead to mistakes.

We hope that these questions from our book will help you improve your preparation and achieve your desired goals.