Zeroes of Quadratic Polynomial
A quadratic polynomial is an algebraic expression of the form ax^2 + bx + c, where a, b, and c are real numbers and a is not equal to 0. The zeroes of a quadratic polynomial are the values of x for which the polynomial equals zero. Geometrically, these zeroes represent the x-coordinates of the points where the parabola (the graph of the quadratic polynomial) intersects the x-axis. Finding the zeroes of quadratic polynomials is one of the most fundamental skills in algebra and is a core topic in the CBSE Class 10 Mathematics curriculum under the chapter on Polynomials. This concept connects algebra with geometry, as the number and nature of zeroes directly correspond to how the parabola interacts with the x-axis. A quadratic polynomial can have at most two zeroes, which aligns with the general principle that a polynomial of degree n has at most n zeroes.
What is Zeroes of Quadratic Polynomial - Definition, Methods, Examples & FAQs?
The zeroes of a quadratic polynomial p(x) = ax^2 + bx + c are the values of x that satisfy the equation p(x) = 0, i.e., the solutions of ax^2 + bx + c = 0. These are also called the roots of the corresponding quadratic equation.
Formal Definition: A real number k is called a zero of the polynomial p(x) if p(k) = 0. For a quadratic polynomial p(x) = ax^2 + bx + c, the value k is a zero if and only if ak^2 + bk + c = 0.
Geometrical Meaning: The graph of a quadratic polynomial y = ax^2 + bx + c is a parabola. If a > 0, the parabola opens upwards (U-shaped). If a < 0, the parabola opens downwards (inverted U-shaped). The zeroes of the polynomial correspond to the x-coordinates of the points where the parabola crosses or touches the x-axis.
Three Possible Cases:
(i) Two distinct zeroes: The parabola intersects the x-axis at two distinct points. This happens when the discriminant D = b^2 - 4ac > 0. The two zeroes are different real numbers.
(ii) One repeated zero (two equal zeroes): The parabola touches the x-axis at exactly one point (the vertex lies on the x-axis). This happens when D = b^2 - 4ac = 0. The polynomial has two equal zeroes, sometimes called a repeated root or a double root.
(iii) No real zeroes: The parabola does not intersect or touch the x-axis at all. It lies entirely above the x-axis (when a > 0) or entirely below (when a < 0). This happens when D = b^2 - 4ac < 0.
The Discriminant: The expression D = b^2 - 4ac is called the discriminant of the quadratic polynomial. It determines the nature and number of zeroes:
D > 0: Two distinct real zeroes
D = 0: Two equal real zeroes (one repeated zero)
D < 0: No real zeroes (zeroes are complex numbers, beyond the Class 10 scope)
Maximum Number of Zeroes: A polynomial of degree n has at most n zeroes. Since a quadratic polynomial has degree 2, it can have at most 2 zeroes. It may have exactly 2, exactly 1 (repeated), or 0 real zeroes.
Zeroes of Quadratic Polynomial Formula
The zeroes of the quadratic polynomial ax^2 + bx + c are found using the Quadratic Formula:
x = (-b +/- sqrt(b^2 - 4ac)) / (2a)
This gives two values:
x1 = (-b + sqrt(b^2 - 4ac)) / (2a)
x2 = (-b - sqrt(b^2 - 4ac)) / (2a)
Where:
a = coefficient of x^2 (must not be zero)
b = coefficient of x
c = constant term
D = b^2 - 4ac is the discriminant
Alternative Methods to Find Zeroes:
1. Factorisation Method (Splitting the Middle Term):
Write bx as the sum of two terms whose product equals ac * x^2.
ax^2 + bx + c = ax^2 + px + qx + c, where p + q = b and p * q = a * c
2. Completing the Square Method:
Rewrite ax^2 + bx + c = 0 as a(x + b/(2a))^2 + (c - b^2/(4a)) = 0 and solve.
3. Graphical Method:
Plot y = ax^2 + bx + c and find where the graph intersects the x-axis.
Derivation and Proof
Let us derive the quadratic formula by applying the method of completing the square to the general quadratic equation ax^2 + bx + c = 0.
Step 1: Start with the general equation
ax^2 + bx + c = 0, where a is not equal to 0.
Step 2: Divide both sides by a
x^2 + (b/a)x + c/a = 0
This is valid because a is not zero.
Step 3: Move the constant term to the right side
x^2 + (b/a)x = -c/a
Step 4: Complete the square on the left side
To complete the square, add the square of half the coefficient of x to both sides.
Half of b/a is b/(2a). Its square is b^2/(4a^2).
x^2 + (b/a)x + b^2/(4a^2) = -c/a + b^2/(4a^2)
Step 5: Write the left side as a perfect square
(x + b/(2a))^2 = b^2/(4a^2) - c/a
Taking LCM on the right side:
(x + b/(2a))^2 = (b^2 - 4ac) / (4a^2)
Step 6: Take the square root of both sides
x + b/(2a) = +/- sqrt(b^2 - 4ac) / (2a)
Note: sqrt(4a^2) = 2|a| = 2a if a > 0, or -2a if a < 0. In either case, the +/- sign absorbs this, so we write 2a in the denominator.
Step 7: Solve for x
x = -b/(2a) +/- sqrt(b^2 - 4ac) / (2a)
x = (-b +/- sqrt(b^2 - 4ac)) / (2a)
This is the quadratic formula, also known as the Sridharacharya formula, named after the 9th-century Indian mathematician Sridharacharya who first described this method.
Verification with a specific polynomial:
Let p(x) = x^2 - 5x + 6. Here a = 1, b = -5, c = 6.
D = (-5)^2 - 4(1)(6) = 25 - 24 = 1
x = (5 +/- sqrt(1)) / 2 = (5 +/- 1) / 2
x1 = (5 + 1)/2 = 3
x2 = (5 - 1)/2 = 2
Verification: p(3) = 9 - 15 + 6 = 0 and p(2) = 4 - 10 + 6 = 0. Both values are indeed zeroes.
Why the Discriminant Matters:
From the formula x = (-b +/- sqrt(D)) / (2a), we see that the square root of D must be real for real zeroes. If D > 0, sqrt(D) is a positive real number, giving two distinct values. If D = 0, sqrt(D) = 0, giving both values equal to -b/(2a). If D < 0, sqrt(D) is not a real number, so no real zeroes exist.
Types and Properties
Zeroes of quadratic polynomials can be classified based on the nature of the discriminant and the methods used to find them.
1. Two Distinct Real Zeroes (D > 0):
When b^2 - 4ac > 0, the polynomial has two different real zeroes. The parabola cuts the x-axis at two distinct points. Example: p(x) = x^2 - 7x + 12 has D = 49 - 48 = 1 > 0, zeroes are x = 3 and x = 4.
2. Two Equal Real Zeroes (D = 0):
When b^2 - 4ac = 0, both zeroes are equal, each being -b/(2a). The parabola just touches the x-axis at its vertex. Example: p(x) = x^2 - 6x + 9 has D = 36 - 36 = 0, both zeroes equal x = 3. Notice that x^2 - 6x + 9 = (x - 3)^2.
3. No Real Zeroes (D < 0):
When b^2 - 4ac < 0, the polynomial has no real zeroes. The parabola lies entirely above or below the x-axis. Example: p(x) = x^2 + x + 1 has D = 1 - 4 = -3 < 0, so it has no real zeroes.
4. Rational Zeroes:
When D is a perfect square, the zeroes are rational numbers. These polynomials can always be factored using the splitting the middle term method. Example: p(x) = 2x^2 + 7x + 3, where D = 49 - 24 = 25. Since sqrt(25) = 5, zeroes are x = (-7 + 5)/4 = -1/2 and x = (-7 - 5)/4 = -3.
5. Irrational Zeroes:
When D > 0 but is not a perfect square, the zeroes are irrational and always occur in conjugate pairs (a + sqrt(b) and a - sqrt(b)). Example: p(x) = x^2 - 2x - 1, where D = 4 + 4 = 8. Zeroes are x = (2 +/- sqrt(8))/2 = 1 +/- sqrt(2).
Methods to Find Zeroes:
(a) Splitting the Middle Term: Best when D is a perfect square and the polynomial can be factored easily with integer or simple rational coefficients.
(b) Quadratic Formula: Works for every quadratic polynomial, regardless of whether it can be factored easily. This is the universal method.
(c) Graphical Method: Plot the polynomial and read off the x-intercepts. This gives approximate values and is useful for visualization.
Solved Examples
Example 1: Example 1: Finding Zeroes by Splitting the Middle Term
Problem: Find the zeroes of p(x) = x^2 - 7x + 10.
Solution:
We need two numbers whose sum is -7 and whose product is 10.
The numbers are -5 and -2, because (-5) + (-2) = -7 and (-5) x (-2) = 10.
x^2 - 7x + 10 = x^2 - 5x - 2x + 10
= x(x - 5) - 2(x - 5)
= (x - 5)(x - 2)
Setting p(x) = 0: (x - 5)(x - 2) = 0
x = 5 or x = 2
Answer: The zeroes are 2 and 5.
Example 2: Example 2: Using the Quadratic Formula
Problem: Find the zeroes of p(x) = 2x^2 - 3x - 5.
Solution:
Here a = 2, b = -3, c = -5.
D = b^2 - 4ac = (-3)^2 - 4(2)(-5) = 9 + 40 = 49
Since D > 0, the polynomial has two distinct real zeroes.
x = (-b +/- sqrt(D)) / (2a) = (3 +/- sqrt(49)) / (2 x 2) = (3 +/- 7) / 4
x1 = (3 + 7)/4 = 10/4 = 5/2
x2 = (3 - 7)/4 = -4/4 = -1
Verification: p(5/2) = 2(25/4) - 3(5/2) - 5 = 25/2 - 15/2 - 10/2 = 0
p(-1) = 2(1) - 3(-1) - 5 = 2 + 3 - 5 = 0
Answer: The zeroes are 5/2 and -1.
Example 3: Example 3: Polynomial with Equal Zeroes
Problem: Find the zeroes of p(x) = 4x^2 - 12x + 9.
Solution:
D = (-12)^2 - 4(4)(9) = 144 - 144 = 0
Since D = 0, the polynomial has two equal zeroes.
x = -b / (2a) = 12 / (2 x 4) = 12/8 = 3/2
Alternatively, factoring: 4x^2 - 12x + 9 = (2x - 3)^2
Setting (2x - 3)^2 = 0: 2x - 3 = 0, so x = 3/2.
Answer: The zeroes are 3/2 and 3/2 (repeated zero).
Example 4: Example 4: Polynomial with No Real Zeroes
Problem: Show that p(x) = 2x^2 + 3x + 4 has no real zeroes.
Solution:
Here a = 2, b = 3, c = 4.
D = b^2 - 4ac = (3)^2 - 4(2)(4) = 9 - 32 = -23
Since D = -23 < 0, the polynomial has no real zeroes.
Geometrical interpretation: The parabola y = 2x^2 + 3x + 4 opens upwards (a = 2 > 0) and its vertex lies above the x-axis. Therefore, the parabola never intersects the x-axis.
The minimum value of the polynomial occurs at x = -b/(2a) = -3/4.
p(-3/4) = 2(9/16) + 3(-3/4) + 4 = 9/8 - 9/4 + 4 = 9/8 - 18/8 + 32/8 = 23/8 > 0
Since the minimum value 23/8 is positive, the polynomial is always positive, confirming no real zeroes.
Example 5: Example 5: Finding Zeroes of a Polynomial with Irrational Roots
Problem: Find the zeroes of p(x) = x^2 + 4x + 1.
Solution:
a = 1, b = 4, c = 1.
D = 16 - 4 = 12 > 0 (but 12 is not a perfect square, so zeroes are irrational).
x = (-4 +/- sqrt(12)) / 2 = (-4 +/- 2*sqrt(3)) / 2 = -2 +/- sqrt(3)
x1 = -2 + sqrt(3) (approximately -2 + 1.732 = -0.268)
x2 = -2 - sqrt(3) (approximately -2 - 1.732 = -3.732)
Answer: The zeroes are -2 + sqrt(3) and -2 - sqrt(3).
Example 6: Example 6: Finding Zeroes by Factoring with Common Factor
Problem: Find the zeroes of p(x) = 3x^2 - 12x.
Solution:
Take out the common factor:
3x^2 - 12x = 3x(x - 4)
Setting p(x) = 0: 3x(x - 4) = 0
Either 3x = 0 or x - 4 = 0
x = 0 or x = 4
Verification: p(0) = 0 - 0 = 0. p(4) = 3(16) - 12(4) = 48 - 48 = 0.
Answer: The zeroes are 0 and 4.
Example 7: Example 7: Finding Zeroes When One Zero Is Given
Problem: If one zero of p(x) = 3x^2 + kx - 2 is 2/3, find k and the other zero.
Solution:
Since 2/3 is a zero: p(2/3) = 0
3(2/3)^2 + k(2/3) - 2 = 0
3(4/9) + 2k/3 - 2 = 0
4/3 + 2k/3 - 2 = 0
(4 + 2k - 6)/3 = 0
2k - 2 = 0
k = 1
So p(x) = 3x^2 + x - 2.
Factoring: 3x^2 + 3x - 2x - 2 = 3x(x + 1) - 2(x + 1) = (3x - 2)(x + 1)
Zeroes: x = 2/3 and x = -1.
Answer: k = 1 and the other zero is -1.
Example 8: Example 8: Word Problem on Zeroes
Problem: The product of two consecutive positive integers is 306. Represent this as a quadratic polynomial and find the integers by determining the zeroes.
Solution:
Let the two consecutive integers be x and x + 1.
Given: x(x + 1) = 306
x^2 + x - 306 = 0
We need two numbers whose product is -306 and sum is 1.
306 = 2 x 153 = 2 x 9 x 17 = 18 x 17
So we need 18 and -17: 18 + (-17) = 1 and 18 x (-17) = -306.
x^2 + 18x - 17x - 306 = 0
x(x + 18) - 17(x + 18) = 0
(x - 17)(x + 18) = 0
x = 17 or x = -18
Since we need positive integers, x = 17.
The consecutive integers are 17 and 18.
Verification: 17 x 18 = 306.
Answer: The integers are 17 and 18.
Example 9: Example 9: Finding a Quadratic Polynomial from Its Zeroes
Problem: Find a quadratic polynomial whose zeroes are 3 + sqrt(5) and 3 - sqrt(5).
Solution:
Let alpha = 3 + sqrt(5) and beta = 3 - sqrt(5).
Sum of zeroes = alpha + beta = (3 + sqrt(5)) + (3 - sqrt(5)) = 6
Product of zeroes = alpha x beta = (3 + sqrt(5))(3 - sqrt(5)) = 9 - 5 = 4
A quadratic polynomial with zeroes alpha and beta is:
p(x) = x^2 - (sum of zeroes)x + (product of zeroes)
p(x) = x^2 - 6x + 4
Verification using the quadratic formula:
x = (6 +/- sqrt(36 - 16))/2 = (6 +/- sqrt(20))/2 = (6 +/- 2*sqrt(5))/2 = 3 +/- sqrt(5). Correct!
Answer: The polynomial is x^2 - 6x + 4.
Example 10: Example 10: Determining the Nature of Zeroes Without Solving
Problem: Without finding the actual zeroes, determine the nature of zeroes of the following polynomials: (i) 3x^2 - 4x + 1, (ii) x^2 + 6x + 9, (iii) 5x^2 - 2x + 3.
Solution:
(i) p(x) = 3x^2 - 4x + 1
D = (-4)^2 - 4(3)(1) = 16 - 12 = 4 > 0
Since D > 0, the polynomial has two distinct real zeroes.
(ii) p(x) = x^2 + 6x + 9
D = 6^2 - 4(1)(9) = 36 - 36 = 0
Since D = 0, the polynomial has two equal real zeroes.
(iii) p(x) = 5x^2 - 2x + 3
D = (-2)^2 - 4(5)(3) = 4 - 60 = -56 < 0
Since D < 0, the polynomial has no real zeroes.
Real-World Applications
Zeroes of quadratic polynomials are foundational to many areas of mathematics, science, and engineering.
Physics - Projectile Motion: When an object is thrown upward, its height h at time t is given by a quadratic equation h = -gt^2/2 + ut + h0. The zeroes of this equation give the times when the object is at ground level, helping determine the total time of flight and the landing point.
Economics - Break-Even Analysis: A company's profit P as a function of units sold x often forms a quadratic equation. The zeroes of this polynomial represent the break-even points where the company transitions from loss to profit or vice versa.
Engineering - Structural Design: Parabolic arches and suspension cables follow quadratic equations. Finding the zeroes determines where the arch meets the ground, which is critical for calculating span length and load distribution.
Geometry - Area Problems: When calculating dimensions of shapes with given area constraints, the resulting equations are often quadratic. The zeroes provide the possible dimensions, with negative values being rejected on physical grounds.
Computer Science - Optimization: Quadratic functions appear in optimization algorithms. Their zeroes and vertex help determine minimum or maximum values, which are used in machine learning loss functions and signal processing.
Daily Life - Financial Planning: Compound interest and loan amortization calculations sometimes involve quadratic equations. The zeroes help find the time period needed to achieve a financial goal or pay off a debt.
Key Points to Remember
- A quadratic polynomial has the form p(x) = ax^2 + bx + c where a is not equal to 0, and it can have at most 2 zeroes.
- The zeroes are the values of x for which p(x) = 0, and geometrically they are the x-coordinates where the parabola crosses the x-axis.
- The discriminant D = b^2 - 4ac determines the nature of zeroes: D > 0 gives two distinct real zeroes, D = 0 gives two equal zeroes, D < 0 gives no real zeroes.
- The quadratic formula x = (-b +/- sqrt(b^2 - 4ac)) / (2a) works for finding zeroes of any quadratic polynomial.
- The splitting the middle term method works best when the discriminant is a perfect square and coefficients are manageable integers.
- If alpha and beta are the zeroes, the polynomial can be written as a(x - alpha)(x - beta).
- Irrational zeroes always occur in conjugate pairs: if (p + sqrt(q)) is a zero, then (p - sqrt(q)) is also a zero.
- The graph of a quadratic polynomial is a parabola. It opens upward if a > 0 and downward if a < 0.
- Always verify your zeroes by substituting them back into the original polynomial to confirm p(zero) = 0.
- CBSE board exams frequently test zeroes through factorisation, quadratic formula application, and word problems.
Practice Problems
- Find the zeroes of the polynomial p(x) = x^2 - 3x - 28 by factorisation.
- Using the quadratic formula, find the zeroes of 3x^2 - 11x + 6.
- Determine the nature of zeroes of 2x^2 + 5x + 4 without actually finding them.
- If the zeroes of x^2 + px + 12 are in the ratio 1:3, find the value of p.
- Find a quadratic polynomial whose zeroes are -3 and 1/2.
- The sum of a number and its reciprocal is 10/3. Form the quadratic polynomial and find the number.
- For what value of k does the polynomial kx^2 - 5x + 3 have two equal zeroes?
- Find the zeroes of 6x^2 - 3 - 7x and verify the relationship between zeroes and coefficients.
Frequently Asked Questions
Q1. What are the zeroes of a quadratic polynomial?
The zeroes of a quadratic polynomial p(x) = ax^2 + bx + c are the values of x for which p(x) equals zero. They are the solutions of the equation ax^2 + bx + c = 0. Graphically, they represent the x-coordinates where the parabola intersects the x-axis.
Q2. How many zeroes can a quadratic polynomial have?
A quadratic polynomial can have at most 2 real zeroes. It may have exactly 2 distinct real zeroes (when D > 0), exactly 1 repeated real zero (when D = 0), or no real zeroes (when D < 0). The maximum number of zeroes equals the degree of the polynomial.
Q3. What is the quadratic formula?
The quadratic formula is x = (-b +/- sqrt(b^2 - 4ac)) / (2a), which gives the zeroes of any quadratic polynomial ax^2 + bx + c. It is derived by completing the square on the general quadratic equation and works universally for all quadratic polynomials.
Q4. What is the discriminant and how does it help?
The discriminant is D = b^2 - 4ac. It tells us the nature of zeroes without solving the equation: if D > 0, there are two distinct real zeroes; if D = 0, there are two equal real zeroes; if D < 0, there are no real zeroes. It also tells whether the zeroes are rational (D is a perfect square) or irrational (D is positive but not a perfect square).
Q5. What is the difference between zeroes and roots?
In the context of Class 10 Mathematics, zeroes and roots refer to the same values. 'Zeroes' is typically used when referring to a polynomial (values where the polynomial is zero), while 'roots' is used when referring to an equation (solutions of the equation). For p(x) = x^2 - 5x + 6, the zeroes of the polynomial and the roots of x^2 - 5x + 6 = 0 are both x = 2 and x = 3.
Q6. How do you find zeroes by splitting the middle term?
To find zeroes by splitting the middle term of ax^2 + bx + c: (1) Find two numbers whose sum equals b and whose product equals a*c. (2) Rewrite the middle term bx using these two numbers. (3) Group the terms in pairs and factor out common factors. (4) Write as a product of two binomials. (5) Set each factor to zero to find the zeroes.
Q7. Can a quadratic polynomial have only one zero?
A quadratic polynomial can have one repeated zero (when D = 0), but technically it still has two zeroes that happen to be equal. For example, x^2 - 4x + 4 = (x - 2)^2 has the zero x = 2 with multiplicity 2. In some contexts, we say it has 'one distinct zero' while acknowledging it has 'two equal zeroes'.
Q8. What does the graph of a quadratic polynomial look like?
The graph of a quadratic polynomial is a parabola (U-shaped curve). If the coefficient of x^2 is positive, the parabola opens upward; if negative, it opens downward. The lowest or highest point is called the vertex. The parabola is symmetric about a vertical line through the vertex. The x-intercepts of the parabola (if any) give the zeroes of the polynomial.
Q9. Why is the coefficient of x^2 not allowed to be zero?
If the coefficient of x^2 (that is, 'a') is zero, the expression becomes bx + c, which is a linear polynomial, not a quadratic polynomial. A quadratic polynomial must have degree exactly 2, which requires the coefficient of x^2 to be non-zero. Linear polynomials have at most 1 zero, not 2.
Q10. How are zeroes of a quadratic polynomial asked in CBSE board exams?
CBSE Class 10 board exams typically ask: (1) Find the zeroes of a given quadratic polynomial and verify the relationship between zeroes and coefficients (2-3 mark questions). (2) Find a quadratic polynomial given its zeroes. (3) Word problems that lead to quadratic equations. (4) Determine the nature of zeroes using the discriminant. Practice all four types for thorough preparation.
Related Topics
- Relationship Between Zeroes and Coefficients
- Graph of Quadratic Polynomial
- Zeroes of a Polynomial
- Sum and Product of Zeroes
- Polynomials in One Variable
- Degree of a Polynomial
- Types of Polynomials
- Value of a Polynomial
- Remainder Theorem
- Factor Theorem
- Factorisation of Polynomials
- Algebraic Identities (Extended)
- (a + b)³ and (a - b)³ Identities
- a³ + b³ and a³ - b³ Identities










