NCERT Solutions Class 10 Maths Chapter 4 Quadratic Equations

Quadratic Equations – Class 10 Maths (Chapter 4) is an important topic for board exams and competitive exam preparation. Students often search for quadratic equation class 10 NCERT solutions to understand concepts clearly and practise different types of problems. NCERT solutions for Class 10 Maths Chapter 4 explain each method step by step, making it easier for students to solve equations confidently.

Class 10 Maths Chapter 4, also known as Chapter 4 Maths Class 10, covers key methods such as factorisation, completing the square, and using the quadratic formula. The quadratic equation class 10 questions answers help students learn how to apply these methods correctly and avoid common mistakes. Clear explanations improve accuracy and speed in exams.

The Class 10 Maths Chapter 4 questions answers are written according to the CBSE syllabus and exam pattern. They include solved examples, exercise questions, and important problems for revision. Regular practice of these solutions helps students strengthen their basics, understand problem-solving steps, and gain confidence. With proper preparation using these NCERT-based solutions, students can score well in Class 10 Maths and build a strong foundation for higher-level mathematics.

Check out: CBSE Class 10 Books

Quadratic Equation Class 10 Questions Answers

Class 10 Maths Chapter 4 Exercise 4.1

1. Check whether the following are quadratic equations:

(i) (x + 1) ² = 2(x – 3)

(ii) x ² – 2x = (–2) (3 – x)

(iii) (x – 2)(x + 1) = (x – 1)(x + 3)

(iv) (x – 3)(2x +1) = x(x + 5)

(v) (2x – 1)(x – 3) = (x + 5)(x – 1)

(vi) x ² + 3x + 1 = (x – 2) ²

(vii) (x + 2) ³ = 2x (x ² – 1)

(viii) x ³ – 4x ² – x + 1 = (x – 2) ³

Solutions:

(i) Given, (x + 1) ² = 2(x – 3) By using the formula for (a+b) ² = a ² +2ab+b ² ⇒ x ² + 2x + 1 = 2x – 6 ⇒ x ² + 7 = 0 The above equation is in the form of ax  ² + bx + c = 0. Therefore, the given equation is a quadratic equation.

(ii) Given, x  ² – 2x = (–2) (3 – x) ⇒ x ² – 2x = -6 + 2x ⇒ x ² – 4x + 6 = 0 The above equation is in the form of ax ² + bx + c = 0. Therefore, the given equation is a quadratic equation.

(iii) Given, (x – 2)(x + 1) = (x – 1)(x + 3) By multiplication ⇒ x² – x – 2 = x ² + 2x – 3 ⇒ 3x – 1 = 0 The above equation is not in the form of ax ² + bx + c = 0. Therefore, the given equation is not a quadratic equation.

(iv) Given, (x – 3)(2x +1) = x(x + 5) By multiplication ⇒ 2x² – 5x – 3 = x ² + 5x ⇒  x ² – 10x – 3 = 0 The above equation is in the form of ax ² + bx + c = 0. Therefore, the given equation is a quadratic equation.

 
(v) Given, (2x – 1)(x – 3) = (x + 5)(x – 1) By multiplication ⇒ 2x² – 7x + 3 = x ² + 4x – 5 ⇒ x ² – 11x + 8 = 0 The above equation is in the form of ax ² + bx + c = 0. Therefore, the given equation is a quadratic equation.

(vi) Given, x  ² + 3x + 1 = (x – 2) ² By using the formula for (a-b) ² =a ² -2ab+b ² ⇒ x ² + 3x + 1 = x ² + 4 – 4x ⇒ 7x – 3 = 0 The above equation is not in the form of ax ² + bx + c = 0. Therefore, the given equation is not a quadratic equation.

(vii) Given, (x + 2)  ³ = 2x(x ² – 1) By using the formula for (a+b) ³ = a ³ +b ³ +3ab(a+b) ⇒ x ³ + 8 + x ² + 12x = 2x ³ – 2x ⇒ x ³ + 14x – 6x ² – 8 = 0 The above equation is not in the form of ax ² + bx + c = 0. Therefore, the given equation is not a quadratic equation.

(viii) Given, x ³ – 4x ² – x + 1 = (x – 2) ³ By using the formula for (a-b) ³ = a ³ -b ³ -3ab(a-b) ⇒  x ³ – 4x ² – x + 1 = x ³ – 8 – 6x ² + 12x ⇒ 2x ² – 13x + 9 = 0 The above equation is in the form of ax ² + bx + c = 0. Therefore, the given equation is a quadratic equation.

2. Represent the following situations in the form of quadratic equations:

(i) The area of a rectangular plot is 528 m ². The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.

(ii) The product of two consecutive positive integers is 306. We need to find the integers.

(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.

(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken

Solutions:

(i) Let us consider, The breadth of the rectangular plot = x m Thus, the length of the plot = (2x + 1) m

As we know, Area of rectangle = length × breadth = 528 m  ² Putting the value of the length and breadth of the plot in the formula, we get (2x + 1) × x = 528 ⇒ 2x ² + x =528 ⇒ 2x ² + x – 528 = 0

Therefore, the length and breadth of the plot satisfy the quadratic equation, 2x  ² + x – 528 = 0, which is the required representation of the problem mathematically.

 
(ii) Let us consider, The first integer number = x Thus, the next consecutive positive integer will be = x + 1 Product of two consecutive integers = x × (x +1) = 306 ⇒ x ² + x = 306 ⇒ x ² + x – 306 = 0 
Therefore, the two integers, x and x+1, satisfy the quadratic equation, x² + x – 306 = 0, which is the required representation of the problem mathematically.

(iii) Let us consider, Age of Rohan’s = x  years Therefore, as per the given question, Rohan’s mother’s age = x + 26 After 3 years, Age of Rohan’s = x + 3 Age of Rohan’s mother will be = x + 26 + 3 = x + 29

The product of their ages after 3 years will be equal to 360, such that (x + 3)(x + 29) = 360 ⇒ x  ² + 29x + 3x + 87 = 360 ⇒ x ² + 32x + 87 – 360 = 0 ⇒ x ² + 32x – 273 = 0

Therefore, the age of Rohan and his mother satisfies the quadratic equation, x  ² + 32x – 273 = 0, which is the required representation of the problem mathematically.

(iv) Let us consider the speed of the train =  x km/h. And the Time taken to travel 480 km = 480/x km/hr. As per second condition, the speed of train = ( x – 8) km/h Also given, the train will take 3 hours to cover the same distance. 

Therefore, time taken to travel 480 km = (480/x)+3 km/h As we know, Speed × Time = Distance Therefore, (  x – 8)(480/ x) + 3 = 480 ⇒ 480 + 3 x – 3840/ x – 24 = 480 ⇒ 3 x – 3840/ x = 24 ⇒ x ² – 8 x – 1280 = 0

Therefore, the speed of the train satisfies the quadratic equation, x ² – 8 x – 1280 = 0, which is the required representation of the problem mathematically.

Read More: NCERT Solutions for Class 10 Maths Chapters 1 Real Number Exercise 1.1

Class 10 Exercise 4.2

1. Find the roots of the following quadratic equations by factorisation:

(i) x ² – 3x – 10 = 0

(ii) 2x  ² + x – 6 = 0

(iii) √2 x ² + 7x + 5√2 = 0

(iv) 2x  ² – x +1/8 = 0

(v) 100x  ² – 20x + 1 = 0

Solutions:

(i) Given, x ² – 3 x – 10 =0 Taking LHS, => x ² – 5 x + 2 x – 10 => x ( x – 5) + 2( x – 5) =>( x – 5)( x + 2) The roots of this equation, x ² – 3 x – 10 = 0 are the values of x for which ( x – 5)( x + 2) = 0 Therefore, x – 5 = 0 or x + 2 = 0 => x = 5 or x = -2

(ii) Given, 2  x ² + x – 6 = 0 Taking LHS, => 2 x ² + 4 x – 3 x – 6 => 2 x ( x + 2) – 3( x + 2) => ( x + 2)(2 x – 3) The roots of this equation, 2 x ² + x – 6=0 are the values of x for which ( x + 2)(2 x – 3) = 0 Therefore, x + 2 = 0 or 2 x – 3 = 0 => x = -2 or x = 3/2

(iii) √2  x ² + 7 x + 5√2=0 Taking LHS, => √2 x ² + 5 x + 2 x + 5√2 => x (√2 x + 5) + √2(√2 x + 5)= (√2 x + 5)( x + √2) The roots of this equation, √2 x ² + 7 x + 5√2=0 are the values of x for which (√2 x + 5)( x + √2) = 0 Therefore, √2 x + 5 = 0 or x + √2 = 0 => x = -5/√2 or x = -√2

(iv) 2  x ² – x +1/8 = 0 Taking LHS, =1/8 (16 x ² – 8 x + 1) = 1/8 (16 x ² – 4 x -4 x + 1) = 1/8 (4 x (4 x – 1) -1(4 x – 1)) = 1/8 (4 x – 1) ² The roots of this equation, 2 x ² – x + 1/8 = 0, are the values of x for which (4 x – 1) ² = 0 Therefore, (4 x – 1) = 0 or (4 x – 1) = 0 ⇒ x = 1/4 or x = 1/4

(v) Given, 100x  ² – 20x + 1=0 Taking LHS, = 100x ² – 10x – 10x + 1 = 10x(10x – 1) -1(10x – 1) = (10x – 1) ² The roots of this equation, 100x ² – 20x + 1=0, are the values of x for which (10x – 1) ² = 0 ∴ (10x – 1) = 0 or (10x – 1) = 0 ⇒x = 1/10 or x = 1/10

2. Solve the problems given in Example 1.

Represent the following situations mathematically:

(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.

(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was 750. We would like to find out the number of toys produced on that day.

Solutions:

(i) Let us say the number of marbles John has = x. Therefore, the number of marbles Jivanti has = 45 – x. 

After losing 5 marbles each, Number of marbles John has =  x – 5 Number of marbles Jivanti has = 45 – x – 5 = 40 – x

Given that the product of their marbles is 124. ∴ ( x – 5)(40 – x ) = 124 ⇒ x ² – 45 x + 324 = 0 ⇒ x ² – 36 x – 9 x + 324 = 0 ⇒ x ( x – 36) -9( x – 36) = 0 ⇒ ( x – 36)( x – 9) = 0 Thus, we can say, x – 36 = 0 or x – 9 = 0 ⇒ x = 36 or x = 9

Therefore, If, John’s marbles = 36, Then, Jivanti’s marbles = 45 – 36 = 9 And if John’s marbles = 9, Then, Jivanti’s marbles = 45 – 9 = 36

(ii) Let us say, number of toys produced in a day be x.

Therefore, cost of production of each toy = Rs(55 –  x )

Given, total cost of production of the toys = Rs 750 ∴  x (55 – x ) = 750 ⇒ x ² – 55 x + 750 = 0 ⇒ x ² – 25 x – 30 x + 750 = 0 ⇒ x ( x – 25) -30( x – 25) = 0 ⇒ ( x – 25)( x – 30) = 0

Thus, either  x -25 = 0 or x – 30 = 0 ⇒ x = 25 or x = 30

Hence, the number of toys produced in a day, will be either 25 or 30.

3. Find two numbers whose sum is 27 and product is 182.

Solution:

Let us say, first number be x and the second number is 27–x.

Therefore, the product of two numbers x(27 – x) = 182 ⇒ x ² – 27x – 182 = 0 ⇒ x ² – 13x – 14x + 182 = 0 ⇒ x(x – 13) -14(x – 13) = 0 ⇒ (x – 13)(x -14) = 0

Thus, either, x = -13 = 0 or x – 14 = 0 ⇒ x = 13 or x = 14 Therefore, if first number = 13, then second number = 27 – 13 = 14 And if first number = 14, then second number = 27 – 14 = 13

 
Hence, the numbers are 13 and 14.

4. Find two consecutive positive integers, sum of whose squares is 365.

Solution:

Let us say the two consecutive positive integers be x and x + 1.

Therefore, as per the given questions,  x ² + ( x + 1) ² = 365 ⇒ x ² + x ² + 1 + 2 x = 365 ⇒ 2 x ² + 2x – 364 = 0 ⇒ x ² + x – 182 = 0 ⇒ x ² + 14 x – 13 x – 182 = 0 ⇒ x ( x + 14) -13( x + 14) = 0 ⇒ ( x + 14)( x – 13) = 0

Thus, either,  x + 14 = 0 or x – 13 = 0, ⇒ x = – 14 or x = 13 since, the integers are positive, so x can be 13, only. ∴ x + 1 = 13 + 1 = 14. Therefore, two consecutive positive integers will be 13 and 14.

5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Solution:

Let us say the base of the right triangle is x cm.

Given, the altitude of right triangle = (x – 7) cm From Pythagoras theorem, we know, Base ² + Altitude ² = Hypotenuse ² ∴ x ² + (x – 7) ² = 13 ² ⇒ x ² + x ² + 49 – 14x = 169 ⇒ 2x ² – 14x – 120 = 0 ⇒ x ² – 7x – 60 = 0 ⇒ x ² – 12x + 5x – 60 = 0 ⇒ x(x – 12) + 5(x – 12) = 0 ⇒ (x – 12)(x + 5) = 0

Thus, either x – 12 = 0 or x + 5 = 0, ⇒ x = 12 or x = – 5

Since sides cannot be negative, x can only be 12. Therefore, the base of the given triangle is 12 cm, and the altitude of this triangle will be (12 – 7) cm = 5 cm.

6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs. 90, find the number of articles produced and the cost of each article.

Solution:

Let us say the number of articles produced is x.

Therefore, cost of production of each article = Rs. (2  x + 3)

Given, total cost of production is Rs. 90 ∴  x (2 x + 3) = 90 ⇒ 2 x ² + 3 x – 90 = 0 ⇒ 2 x ² + 15 x -12 x – 90 = 0 ⇒ x (2 x + 15) -6(2 x + 15) = 0 ⇒ (2 x + 15)( x – 6) = 0

Thus, either 2  x + 15 = 0 or x – 6 = 0 ⇒ x = -15/2 or x = 6

As the number of articles produced can only be a positive integer, x can only be 6. Hence, number of articles produced = 6 Cost of each article = 2 × 6 + 3 = Rs. 15.

Read More: NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

Class 10 Maths Chapter 4 Exercise 4.3

1. Find the nature of the roots of the following quadratic equation. If the real roots exist, find them:

(i) 2x2 – 3x + 5 = 0

(ii) 3x 2 -4√3x+4 = 0

(iii) 2x2– 6x + 3 = 0

Answer:

We know that the quadratic equation ax 2 +bx+c = 0 has
(a) Two distinct real roots, if b 2 -4ac > 0,
(b) Two equal real roots, if b 2 -4ac = 0,
(c) No real roots, if b 2 -4ac < 0.

(i) 2x 2 -3x+5 = 0

Answer:
Comparing the given quadratic equation with the general form of quadratic equation ax 2 +bx+c = 0,

we get a = 2, b = -3 and c = 5
Then, b 2 -4ac = (-3) 2 -4×2×5 = 9-40 = -31 < 0
Hence, the given quadratic equation has no real roots.

(ii) 3x 2 -4√3x+4 = 0

Answer:
Comparing the given quadratic equation with the general form of quadratic equation ax 2 +bx+c = 0

we get a = 3, b = -4√3 and c = 4
Then, b 2 -4ac = (-4√3) 2 -4×3×4 = 48-48 = 0
Therefore, real roots exist for the given equation,

 and they are equal to each other.

And the roots will be

 

Therefore, the roots are

 

(iii) 2x 2 -6x+3 = 0

Answer:
Comparing the given quadratic equation with the general form of quadratic equation ax2+bx+c = 0,
we get a = 2, b = -6 and c = 3
then, b 2 -4ac = (-6) 2 -4×2×3 = 36-24 = 12 > 0
Hence, the given quadratic equation has two distinct real roots.

Applying the quadratic formulato find roots,

2. Find the value of k for each of the following quadratic equations, so that they have two equal roots.

(i) 2x 2 +kx+3 = 0

Answer:
Comparing the given quadratic equation with the general form of quadratic equation ax 2 +bx+c = 0, we get
a = 2, b = k and c = 3
Now the given quadratic equation have two equal roots if
b 2 -4ac = 0

(k)2-4×2×3 = 0
k 2 -24 = 0
k 2 = 24
k = ±√24
k = ± 2√6

Therefore, the required value of k is ± 2√6

(ii)  kx(x-2)+6 = 0

Answer:
The given quadratic equation can be written as
kx 2 -2kx+6 = 0….(1)
Comparing the quadratic equation (1) with the general form of quadratic equation ax 2 +bx+c = 0 we get
a = k, b = -2k and c = 6

Now the given quadratic equation have two equal roots if
b 2 -4ac = 0
(-2k) 2 -4×k×6 = 0
4k 2 -24k = 0
4k(k-6) = 0
k(k-6) = 0
k = 0
Or, k-6 = 0
k = 6

If k = 0, then equation will not have x 2 and x, which is not possible because the given equation is quadratic.
Therefore, the required value of k is 6.

3. Is it possible to design a rectangular mango grove whose length is twice its breadth and the area is 800 m 2? If so, find its length and breadth.

Answer:
Let the breadth of the mango grove be x m and the length is 2x m.
Area = length × breadth
= x × 2x
= 2x2 m 2
Then by the given condition,
2x 2 = 800
x 2 = 400
x = ±√400
x = ±20
Since length cannot be negative, then x ≠ -20
Hence, it is possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m 2, and its breadth is 20 m, and length is 20×2 = 40 m


4. 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.

Answer:
Let the age of 1st friend is x.
Then the age of 2nd friend is (20-x)
Four years ago their age was (x-4) and (20-x-4)
Then, using the given condition, we have
(x-4)(16-x) = 48
16x – 64 – x 2 +4x = 48
20x – x 2 – 64 – 48 = 0
x 2 – 20x + 112 = 0……(1)

Now, comparing the above quadratic equation with the general form of quadratic equation ax 2 +bx+c = 0 we get
a = 1, b = -20 and c = 112
then, b 2 -4ac = (-20) 2 -4×1×112 = 400-448 = -48 > 0

Therefore, no real root is possible for this equation, and hence, this situation is not possible.


5. Is it possible to design a rectangular park of perimeter 80 m and area 400 m 2? If so, find its length and breadth.

Answer:
Let the length and breadth of the park be “l” and “b”
Area of rectangle = 2(l + b)
2(l + b) = 80
l + b = 40
b = 40 – l
Then, Area = l(40 – l)
Then by the given condition,
l(40 – l) = 400
40l  – l 2 -400 = 0
l 2 – 40l + 400 = 0….(1)

Now, comparing the above quadratic equation with the general form of quadratic equation ax 2 +bx+c = 0 we get
a = 1, b = -40 and c = 400
Then, b 2 -4ac = (-40) 2 -4×1×400 = 1600-1600 = 0
As the quadratic equation has two equal roots, the given situation is possible.

Therefore, length of park, l = 20 m

And breadth of park, b = 40 - l = 40 - 20 = 20 m.

Read More: NCERT Solutions for Class 10 Maths Chapter 3

Identification and Formulation of Quadratic Equations

The first step in chapter 4 maths class 10 is learning how to identify a quadratic equation. Any equation that can be reduced to the standard form ax^2 + bx + c = 0 is a quadratic equation. The most crucial condition is that the highest power of the variable x must be 2, and the coefficient a cannot be zero.

 

In chapter 4 maths class 10, students are often asked to check whether a given mathematical statement represents a quadratic equation. For example, (x-2)^2 + 1 = 2x - 3 is a quadratic equation because, upon expansion and simplification, it retains the x^2 term. Formulating these equations from word problems is a vital skill. Whether it is finding the dimensions of a rectangular plot or determining the speed of a train, the ability to translate a situational problem into a quadratic expression is the hallmark of mathematical proficiency.

Methods of Solving Quadratic Equations

Solving a quadratic equation means finding the "roots" or the values of x that satisfy the equation. The chapter 4 maths class 10 curriculum introduces several key methods to achieve this:

1. Factorisation (Splitting the Middle Term): This method involves finding two numbers that multiply to ac and add up to b. Once the middle term is split, the equation is factorised into two linear expressions. This is the primary focus of chapter 4 maths class 10.

2. Quadratic Formula: For equations that are difficult to factorise, the quadratic formula is a universal tool. The roots are given by:
   
   x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

3. Completing the Square: This method involves transforming the equation into a perfect square trinomial, allowing for easy extraction of the square root.

Check Out: CBSE Class 10 Sample Papers

Nature of Roots and the Discriminant

One of the most theoretically significant parts of chapter 4 maths class 10 is the concept of the Discriminant, denoted by D = b^2 - 4ac. The value of the discriminant determines the nature of the roots without actually solving the equation:

Understanding the discriminant is essential for solving chapter 4 maths class 10 extra questions, as it helps students predict the behavior of the quadratic graph (parabola) and its intersection with the x-axis.

Application of Quadratic Equations in Real Life

Quadratic equations are not just abstract symbols; they are tools for solving practical challenges. The source material highlights several applications:

• Geometry: Finding the length and breadth of shapes when the area and a relationship between dimensions are known.

• Motion: Calculating the time taken for an object to reach a certain height or the speed of a boat in upstream and downstream conditions.

• Business: Determining the cost of production and the number of articles produced in a day.

By mastering these applications, students move from rote learning to practical problem-solving, which is a key objective of the NCERT curriculum.

Also Check, CBSE Class 10 Previous Year Papers

Original Framing: The "Parabolic Balance" Perspective

A unique way to view the concepts in chapter 4 maths class 10 is through the lens of Parabolic Balance. Often, we see the roots of an equation as just two numbers. However, this framing suggests that every quadratic equation represents a symmetrical path—a parabola. The roots are the "grounding points" where this path meets reality (the x-axis). The discriminant D acts as a "height adjustment" tool; it tells us if the curve is low enough to touch the ground twice, perfectly balanced to touch it once, or floating entirely in the air. This perspective helps students visualize algebra as a geometric dance, making the "nature of roots" a physical reality rather than a memorized rule.

Benefits of PW CBSE Class 10 Maths Study Material

Clear Concept Explanation

PW CBSE Class 10 Maths Study Material explains every topic in simple and easy language. Step-by-step solutions help students understand concepts clearly and reduce confusion in difficult chapters.

NCERT and CBSE Aligned

The study material is strictly based on the latest CBSE syllabus and NCERT textbooks. This ensures students study only exam-relevant topics and stay focused on what matters most.

Plenty of Practice Questions

It includes solved examples, exercise questions, and previous years’ questions. Regular practice helps improve accuracy, speed, and problem-solving skills.

Effective Revision Support

Chapter-wise summaries, important formulas, and quick revision notes make last-minute revision easier and stress-free.

Builds Confidence for Exams

With structured content and regular practice, PW CBSE Class 10 Maths Study Material helps students feel confident and perform better in board exams.

NCERT Solutions Class 10 Chapter 4 FAQs

1. How can I tell if a situation can be represented by a quadratic equation?

If the relationship between the variables involves the square of one variable (like Area = Length \times Breadth where one is a function of the other), it can usually be represented as a quadratic equation.

2. What is the most common mistake in exercise 4.2?

The most common mistake is incorrect sign placement while splitting the middle term. Always ensure that the product of the two numbers equals a times c, including the positive or negative signs.

3. Is it possible for a quadratic equation to have only one root?

Mathematically, a quadratic equation always has two roots. However, if the discriminant D = 0, those two roots are identical, which is why we often say it has "one unique real root."

4. Why is the quadratic formula preferred over factorisation?

While factorisation is faster for simple numbers, the quadratic formula is a universal method that works even when the roots are irrational or decimal values that are impossible to find by splitting the middle term.

5. What does a negative discriminant imply in a real-world problem?

If you are solving a physical problem (like finding the time an object takes to reach a certain height) and you get D < 0, it often means that the situation described is physically impossible under the given conditions.