CBSE Class 10 Mathematics (2026–27)
Chapter 4: Quadratic Equations
20 Important Questions and Answers
Q1. What is a quadratic equation? Give its standard form.
Answer:
A quadratic equation is an equation of degree two in one variable. It contains the square of the variable as its highest power. The standard form of a quadratic equation is:
[ax^2 + bx + c = 0]
where (a), (b), and (c) are real numbers and (a \neq 0). If (a = 0), the equation becomes linear and is no longer quadratic. Examples of quadratic equations are (x^2 + 5x + 6 = 0) and (2x^2 – 7x + 3 = 0). Quadratic equations are important because they appear in various mathematical and real-life situations involving area, motion, and optimization problems.
Q2. How can a quadratic equation be solved by factorization?
Answer:
The factorization method involves expressing the quadratic equation as a product of two linear factors. First, write the equation in standard form (ax^2 + bx + c = 0). Then find two numbers whose product equals (ac) and whose sum equals (b). Split the middle term using these numbers and factor by grouping. After factorization, set each factor equal to zero and solve for the variable. For example, (x^2 + 5x + 6 = 0) can be written as ((x + 2)(x + 3) = 0). Therefore, the solutions are (x = -2) and (x = -3). This method is simple when the quadratic expression can be easily factorized.
Q3. Solve (x^2 – 5x + 6 = 0) by factorization.
Answer:
To solve (x^2 – 5x + 6 = 0), we use factorization. We need two numbers whose product is (6) and sum is (-5). These numbers are (-2) and (-3).
So,
[x^2 – 5x + 6 = x^2 – 2x – 3x + 6]
Grouping terms:
[x(x – 2) – 3(x – 2)]
[(x – 2)(x – 3) = 0]
Using the zero-product property:
[x – 2 = 0 \quad \text{or} \quad x – 3 = 0]
Thus,
[x = 2 \quad \text{or} \quad x = 3]
Hence, the roots of the equation are 2 and 3. Both values satisfy the original equation.
Q4. Explain the method of completing the square.
Answer:
Completing the square is a method used to solve quadratic equations by converting them into a perfect square form. First, move the constant term to the right side. Then divide by the coefficient of (x^2) if necessary. Add the square of half the coefficient of (x) to both sides. This creates a perfect square trinomial on the left side. After taking square roots on both sides, solve for the variable. This method is useful because it leads to the derivation of the quadratic formula. It also helps in understanding the structure of quadratic equations and their graphical representation as parabolas.
Q5. Derive the quadratic formula.
Answer:
Consider the quadratic equation:
[ax^2 + bx + c = 0]
Divide by (a):
[x^2 + \frac{b}{a}x + \frac{c}{a}=0]
Move the constant term:
[x^2 + \frac{b}{a}x = -\frac{c}{a}]
Add (\left(\frac{b}{2a}\right)^2) to both sides:
[\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}]
Taking square roots:
[x+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a}]
Therefore,
[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}]
This formula is known as the quadratic formula and can solve any quadratic equation.
Q6. What is the discriminant of a quadratic equation?
Answer:
The discriminant of a quadratic equation is the expression (b^2 – 4ac) in the quadratic formula. It helps determine the nature of the roots without solving the equation completely. The discriminant is usually represented by (D).
[D = b^2 – 4ac]
If (D > 0), the equation has two distinct real roots. If (D = 0), the equation has two equal real roots. If (D < 0), the equation has no real roots and the roots are imaginary. Thus, the discriminant provides valuable information about the solutions of a quadratic equation and helps in analyzing the equation quickly.
Q7. How does the discriminant determine the nature of roots?
Answer:
The nature of roots of a quadratic equation depends on the value of its discriminant (D = b^2 – 4ac).
- If (D > 0), there are two distinct real roots.
- If (D = 0), there are two equal real roots.
- If (D < 0), there are no real roots.
For example, in (x^2 – 5x + 6 = 0), (D = 25 – 24 = 1), which is positive, so the roots are real and distinct. This concept helps students determine the type of solutions without actually solving the equation. The discriminant is therefore an important tool in quadratic equations.
Q8. Find the roots of (x^2 – 7x + 12 = 0).
Answer:
Given:
[x^2 – 7x + 12 = 0]
We need two numbers whose product is 12 and sum is -7. These numbers are -3 and -4.
Therefore,
[x^2 – 3x – 4x + 12 = 0]
Grouping:
[x(x – 3) – 4(x – 3)]
[(x – 3)(x – 4) = 0]
Hence,
[x – 3 = 0 \quad \text{or} \quad x – 4 = 0]
So,
[x = 3 \quad \text{or} \quad x = 4]
Thus, the roots of the quadratic equation are 3 and 4. Verification shows that both satisfy the original equation.
Q9. Find the roots of (2x^2 – 5x – 3 = 0) using the quadratic formula.
Answer:
Given:
[2x^2 – 5x – 3 = 0]
Here, (a = 2), (b = -5), and (c = -3).
Using the quadratic formula:
[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}]
[x=\frac{5\pm\sqrt{25+24}}{4}]
[x=\frac{5\pm7}{4}]
Therefore,
[x=\frac{12}{4}=3]
or
[x=\frac{-2}{4}=-\frac12]
Hence, the roots are (3) and (-\frac12). The quadratic formula provides a systematic method for solving all quadratic equations.
Q10. What are real and equal roots?
Answer:
Real and equal roots occur when the discriminant of a quadratic equation is zero.
[D = b^2 – 4ac = 0]
In this case, both roots have the same value. For example:
[x^2 – 6x + 9 = 0]
Here,
[D = 36 – 36 = 0]
Using factorization:
[(x – 3)^2 = 0]
Thus, (x = 3) is repeated twice. Such roots are called real and equal roots. Graphically, the parabola touches the x-axis at only one point. This situation is important in many mathematical and practical applications where a repeated solution exists.
Q11. Explain the relationship between roots and coefficients.
Answer:
For a quadratic equation:
[ax^2 + bx + c = 0]
with roots (\alpha) and (\beta),
[\alpha + \beta = -\frac{b}{a}]
and
[\alpha\beta = \frac{c}{a}]
These relationships help find the sum and product of roots directly from the coefficients without solving the equation. They are useful in forming quadratic equations when roots are known and in verifying solutions. This concept establishes a strong connection between the algebraic form of the equation and its roots, making problem-solving more efficient.
Q12. Form a quadratic equation whose roots are 2 and 5.
Answer:
Let the roots be 2 and 5.
The quadratic equation with roots (\alpha) and (\beta) is:
[x^2 – (\alpha+\beta)x + \alpha\beta = 0]
Here,
[\alpha+\beta = 2+5 = 7]
and
[\alpha\beta = 2 \times 5 = 10]
Substituting:
[x^2 – 7x + 10 = 0]
Therefore, the required quadratic equation is:
[x^2 – 7x + 10 = 0]
Verification by factorization gives ((x-2)(x-5)=0), confirming that the roots are indeed 2 and 5.
Q13. What is the importance of quadratic equations in daily life?
Answer:
Quadratic equations are widely used in real-life situations. They help calculate areas of fields, gardens, and rooms. In physics, they are used to study the motion of objects, such as projectiles. Engineers use quadratic equations in designing structures, bridges, and machines. Economists use them to analyze profit and cost functions. Architects apply them while designing arches and curved structures. Quadratic equations also appear in sports, where the path of a ball often forms a parabola. Thus, these equations have practical applications in science, engineering, economics, and everyday problem-solving, making them an important part of mathematics.
Q14. Find the discriminant of (3x^2 + 2x – 1 = 0) and state the nature of roots.
Answer:
Given:
[3x^2 + 2x – 1 = 0]
Here,
[a=3,\quad b=2,\quad c=-1]
The discriminant is:
[D=b^2-4ac]
[D=(2)^2-4(3)(-1)]
[D=4+12=16]
Since (D = 16 > 0), the roots are real and distinct. Moreover, because 16 is a perfect square, the roots are rational numbers. Therefore, the equation has two different real rational roots. The discriminant provides this information without solving the equation completely.
Q15. Solve (x^2 + 2x – 15 = 0).
Answer:
We need two numbers whose product is (-15) and sum is (2). These numbers are (5) and (-3).
Therefore,
[x^2 + 5x – 3x – 15 = 0]
Grouping:
[x(x + 5) – 3(x + 5)]
[(x + 5)(x – 3)=0]
Hence,
[x + 5 = 0]
or
[x – 3 = 0]
Thus,
[x = -5]
or
[x = 3]
Therefore, the roots of the quadratic equation are (-5) and (3). Both values satisfy the original equation when substituted.
Q16. What happens when the discriminant is negative?
Answer:
When the discriminant (D = b^2 – 4ac) is negative, the quadratic equation has no real roots. This is because the square root of a negative number is not defined in the set of real numbers. For example:
[x^2 + x + 1 = 0]
Here,
[D = 1 – 4 = -3]
Since (D < 0), the equation has no real solutions. The roots are imaginary or complex. Graphically, the parabola does not intersect the x-axis. Understanding negative discriminants helps students identify equations that do not have real-number solutions.
Q17. Why must the coefficient (a) not be zero in a quadratic equation?
Answer:
In the standard form (ax^2 + bx + c = 0), the coefficient (a) must not be zero because the equation would lose its quadratic nature. If (a = 0), the term (ax^2) disappears and the equation becomes:
[bx + c = 0]
which is a linear equation of degree one. A quadratic equation is defined by having the highest power of the variable equal to 2. Therefore, (a \neq 0) is an essential condition. Without this condition, methods such as factorization, completing the square, and the quadratic formula would no longer apply.
Q18. Solve (x^2 – 9 = 0).
Answer:
Given:
[x^2 – 9 = 0]
This is a difference of squares.
[x^2 – 3^2 = 0]
Therefore,
[(x – 3)(x + 3)=0]
Using the zero-product property:
[x – 3 = 0]
or
[x + 3 = 0]
Hence,
[x = 3]
or
[x = -3]
Thus, the roots of the equation are 3 and -3. These roots are real and distinct. Verification confirms that substituting either value into the original equation gives zero.
Q19. Explain how to verify the roots of a quadratic equation.
Answer:
To verify the roots of a quadratic equation, substitute the obtained values into the original equation. If the left-hand side becomes zero, the value is a correct root. For example, consider (x^2 – 5x + 6 = 0) with roots 2 and 3. Substituting (x = 2):
[2^2 – 5(2) + 6 = 4 – 10 + 6 = 0]
Substituting (x = 3):
[3^2 – 5(3) + 6 = 9 – 15 + 6 = 0]
Since both values satisfy the equation, they are verified roots. Verification ensures accuracy and helps detect calculation mistakes.
Q20. Compare factorization and quadratic formula methods.
Answer:
Factorization and the quadratic formula are two common methods for solving quadratic equations. Factorization is quicker when the quadratic expression can be easily broken into factors. However, not all equations are easily factorized. The quadratic formula works for every quadratic equation, regardless of whether factorization is possible. Although it involves more calculations, it guarantees the correct roots. Factorization is often preferred for simple equations, while the quadratic formula is useful for complex equations. Both methods are important and should be mastered by students to solve different types of quadratic equations effectively in examinations and practical applications.