Quadratic Equations

Quadratic Equations : Formulas, Theorems & Examples (Class 10 Maths)

1.1 What is a Polynomial?

Before diving into quadratic polynomials, let’s briefly remember what a polynomial is:

A Polynomial is an algebraic expression made up of variables and coefficients, combined using addition, subtraction, and multiplication, where the exponents of the variables are always non-negative integers (whole numbers).

Examples of Polynomials: 3x² + 2x – 5, y³ – 7y + 1, 8 (constant polynomial)

Non-Examples (Not Polynomials): x⁻¹ + 2, √x + 3, x/y + 1 (because of negative exponents, fractional exponents, or variables in the denominator)

1.2 Definition of a Quadratic Polynomial:

A Quadratic Polynomial is a specific type of polynomial. It is a polynomial of degree 2. This means the highest power of the variable in the polynomial is 2.

The standard form of a quadratic polynomial in variable ‘x’ is:

P(x) = ax² + bx + c

Where:

  • ‘a’, ‘b’, and ‘c’ are real numbers. These are the coefficients.
    • ‘a’ is the coefficient of x²
    • ‘b’ is the coefficient of x
    • ‘c’ is the constant term
  • Crucially, ‘a’ must not be equal to 0 (a ≠ 0). If ‘a’ were 0, the x² term would disappear, and the polynomial would become linear (bx + c) or constant (c), not quadratic.

In simpler words: A quadratic polynomial is like an expression with three main parts: something times x-squared, plus something times x, plus a constant number. The most important part is the x-squared term, and it must be there (meaning ‘a’ can’t be zero).

1.3 Examples of Quadratic Polynomials (with Coefficients Identified):

Let’s look at some examples and identify the coefficients ‘a’, ‘b’, and ‘c’ in each case:

Example 1: P(x) = 2x² + 5x – 3

Here, a = 2, b = 5, c = -3

Example 2: Q(x) = -x² + 7x + 10

Here, a = -1 (remember -x² is the same as -1x²), b = 7, c = 10

Example 3: R(x) = x² – 4x

Here, a = 1 (x² is 1x²), b = -4, c = 0 (there’s no constant term, so c is 0)

Example 4: S(x) = 3x² + 9

Here, a = 3, b = 0 (there’s no ‘x’ term, so b is 0), c = 9

Example 5: T(x) = -5x²

Here, a = -5, b = 0, c = 0 (only the x² term is present)

1.4 Non-Examples (Not Quadratic Polynomials):

  • U(x) = 4x + 1 (This is a linear polynomial, degree 1, because the highest power of x is 1. It’s of the form bx + c, where a=0)
  • V(x) = x³ – 2x² + x – 6 (This is a cubic polynomial, degree 3, because the highest power of x is 3. It has an x³ term, so it’s not quadratic)
  • W(x) = 7 (This is a constant polynomial, degree 0, because there’s no variable ‘x’. It’s just a constant term ‘c’, where a=0 and b=0)

2. ZEROS OF A QUADRATIC POLYNOMIAL

2.1 What are Zeros?

A zero of a polynomial P(x) is a value of ‘x’ for which the polynomial becomes equal to zero. In other words, if we substitute a certain value for ‘x’ into P(x) and the result is 0, then that value is a zero of the polynomial.

For a quadratic polynomial P(x) = ax² + bx + c, a zero is a value of ‘x’ that satisfies the equation:

ax² + bx + c = 0

2.2 Finding Zeros Graphically:

We can find the zeros of a quadratic polynomial graphically by looking at its graph. As we learned earlier, the graph of y = P(x) = ax² + bx + c is a parabola.

The zeros of the polynomial are the x-coordinates of the points where the parabola intersects the x-axis. These are the points where y = 0, which is exactly when P(x) = 0.

  • Two Intersections: If the parabola intersects the x-axis at two distinct points, the quadratic polynomial has two distinct real zeros.
  • One Intersection (Vertex on x-axis): If the parabola touches the x-axis at exactly one point (its vertex lies on the x-axis), the quadratic polynomial has one real zero (a repeated zero or equal zeros).
  • No Intersections: If the parabola does not intersect the x-axis at all, the quadratic polynomial has no real zeros. In this case, the zeros are complex numbers (which you will study in higher classes).

quadratic_zeros

2.3 Relationship between Zeros and Coefficients of a Quadratic Polynomial

This is a very important relationship in quadratic polynomials! Let’s say α (alpha) and β (beta) are the two zeros of the quadratic polynomial P(x) = ax² + bx + c. Then, there are direct formulas connecting these zeros to the coefficients a, b, and c:

  • Sum of Zeros (α + β):

α + β = -b/a (Negative of the coefficient of x divided by the coefficient of x²)

  • Product of Zeros (αβ):

αβ = c/a (Constant term divided by the coefficient of x²)

These relationships are extremely useful for:

  • Verifying if you’ve found the zeros correctly.
  • Finding the zeros if you know the sum and product.
  • Constructing a quadratic polynomial if you know its zeros.

Example 6: Verifying the Relationship

Consider the quadratic polynomial P(x) = x² – 5x + 6. We can factorize it as P(x) = (x – 2)(x – 3). Therefore, the zeros are x = 2 and x = 3. Let α = 2 and β = 3.

Let’s verify the relationships:

  • Sum of Zeros (α + β) = 2 + 3 = 5

From the formula: -b/a = -(-5)/1 = 5 (Verified!)

  • Product of Zeros (αβ) = 2 * 3 = 6

From the formula: c/a = 6/1 = 6 (Verified!)

3. FINDING ZEROS OF A QUADRATIC POLYNOMIAL

There are different methods to find the zeros of a quadratic polynomial. We’ll discuss two common methods for Class 10:

3.1 Factorization Method (Splitting the Middle Term)

This method works if the quadratic polynomial can be easily factorized into two linear factors. The steps are:

  1. Write the quadratic polynomial in standard form: ax² + bx + c
  2. Find two numbers whose sum is ‘b’ (the coefficient of x) and whose product is ‘ac’ (the product of the coefficient of x² and the constant term).
  3. Split the middle term ‘bx’ using these two numbers.
  4. Factor by grouping the terms.
  5. Set each factor equal to zero and solve for ‘x’ to get the zeros.

Example 7: Factorization Method

Find the zeros of P(x) = x² – 5x + 6.

  1. Polynomial is already in standard form.
  2. We need two numbers whose sum is -5 and product is (1)(6) = 6. These numbers are -2 and -3 (-2 + (-3) = -5 and (-2) * (-3) = 6).
  3. Split the middle term: x² – 2x – 3x + 6
  4. Factor by grouping: x(x – 2) – 3(x – 2) = (x – 2)(x – 3)
  5. Set factors to zero:
    • x – 2 = 0 => x = 2
    • x – 3 = 0 => x = 3

Therefore, the zeros are 2 and 3.

Example 8: Factorization Method (with a coefficient for x²)

Find the zeros of Q(x) = 2x² + 7x + 3.

  1. Polynomial is in standard form.
  2. We need two numbers whose sum is 7 and product is (2)(3) = 6. These numbers are 6 and 1 (6 + 1 = 7 and 6 * 1 = 6).
  3. Split the middle term: 2x² + 6x + 1x + 3
  4. Factor by grouping: 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)
  5. Set factors to zero:
    • 2x + 1 = 0 => 2x = -1 => x = -1/2
    • x + 3 = 0 => x = -3

Therefore, the zeros are -1/2 and -3.

3.2 Quadratic Formula (Sridhar Acharya Formula)

The quadratic formula is a general formula that can be used to find the zeros of any quadratic polynomial, even if it’s not easily factorizable. For a quadratic polynomial ax² + bx + c, the zeros are given by:

x = [-b ± √(b² – 4ac)] / (2a)

This formula gives you two possible values for ‘x’, corresponding to the two zeros (α and β), using the ‘+’ and ‘-‘ signs respectively.

The part under the square root, (b² – 4ac), is called the Discriminant (D). D = b² – 4ac. The discriminant is very important because it tells us about the nature of the roots (as we’ll see in the next section).

Example 9: Quadratic Formula

Find the zeros of P(x) = x² – 5x + 6 using the quadratic formula.

Here, a = 1, b = -5, c = 6.

x = [-(-5) ± √((-5)² – 4 * 1 * 6)] / (2 * 1)

x = [5 ± √(25 – 24)] / 2

x = [5 ± √1] / 2

x = [5 ± 1] / 2

So, we have two solutions:

  • x = (5 + 1) / 2 = 6 / 2 = 3
  • x = (5 – 1) / 2 = 4 / 2 = 2

The zeros are 2 and 3, which matches what we found by factorization.

Example 10: Quadratic Formula (with irrational roots)

Find the zeros of R(x) = x² – 4x – 1.

Here, a = 1, b = -4, c = -1.

x = [-(-4) ± √((-4)² – 4 * 1 * (-1))] / (2 * 1)

x = [4 ± √(16 + 4)] / 2

x = [4 ± √20] / 2

x = [4 ± √(4 * 5)] / 2

x = [4 ± 2√5] / 2

x = 2 ± √5

So, the zeros are 2 + √5 and 2 – √5. These are irrational zeros, and factorization would have been difficult in this case. The quadratic formula handles it easily.

4. DISCRIMINANT AND NATURE OF ROOTS

4.1 The Discriminant (D = b² – 4ac)

As mentioned earlier, the discriminant (D) of a quadratic polynomial ax² + bx + c is given by:

D = b² – 4ac

The discriminant is a crucial value because it determines the nature of the roots (zeros) of the quadratic polynomial and the quadratic equation ax² + bx + c = 0. There are three cases based on the value of D:

4.2 Nature of Roots based on Discriminant:

Case 1: D > 0 (Discriminant is positive)

  • The quadratic polynomial has two distinct real zeros.
  • Graphically, the parabola intersects the x-axis at two distinct points.
  • In the quadratic formula, √(D) is a real number, and we get two different real values for ‘x’ due to the ± sign.

Case 2: D = 0 (Discriminant is zero)

  • The quadratic polynomial has one real zero (a repeated zero or two equal real zeros).
  • Graphically, the parabola touches the x-axis at exactly one point (the vertex is on the x-axis).
  • In the quadratic formula, √(D) = √0 = 0, so the ± part becomes ± 0, and we get only one value for ‘x’: x = -b / (2a).

Case 3: D < 0 (Discriminant is negative)

  • The quadratic polynomial has no real zeros. The zeros are complex numbers (not real numbers).
  • Graphically, the parabola does not intersect the x-axis at all.
  • In the quadratic formula, √(D) is the square root of a negative number, which is not a real number. Therefore, there are no real solutions.

Example 11: Discriminant and Nature of Roots

Let’s analyze the nature of roots for different quadratic polynomials using the discriminant:

(a) P(x) = x² – 5x + 6

a = 1, b = -5, c = 6

D = b² – 4ac = (-5)² – 4 * 1 * 6 = 25 – 24 = 1

D = 1 > 0. Therefore, two distinct real zeros (which we found to be 2 and 3).

(b) Q(x) = x² – 4x + 4

a = 1, b = -4, c = 4

D = b² – 4ac = (-4)² – 4 * 1 * 4 = 16 – 16 = 0

D = 0. Therefore, one real zero (repeated zero). (You can verify by factorization: x² – 4x + 4 = (x – 2)², so the zero is x = 2, repeated).

(c) R(x) = x² + x + 1

a = 1, b = 1, c = 1

D = b² – 4ac = (1)² – 4 * 1 * 1 = 1 – 4 = -3

D = -3 < 0. Therefore, no real zeros. The zeros are complex numbers.

5. FORMING A QUADRATIC POLYNOMIAL FROM ITS ZEROS

If you are given the zeros of a quadratic polynomial, you can construct the polynomial itself. Let α and β be the zeros. Then, the quadratic polynomial can be written as:

P(x) = k(x – α)(x – β)

Where ‘k’ is any non-zero constant real number. ‘k’ just scales the polynomial but doesn’t change the zeros. For simplicity, we often take k = 1.

Expanding this, we get:

P(x) = k[x² – (α + β)x + αβ]

Notice that (α + β) is the sum of zeros and αβ is the product of zeros. So, if you know the sum and product of zeros, you can directly write the polynomial as:

P(x) = k[x² – (Sum of Zeros)x + (Product of Zeros)]

Example 12: Forming a Quadratic Polynomial

Form a quadratic polynomial whose zeros are 3 and -2.

Let α = 3 and β = -2.

  • Sum of zeros (α + β) = 3 + (-2) = 1
  • Product of zeros (αβ) = 3 * (-2) = -6

Using the formula with k = 1:

P(x) = x² – (Sum of Zeros)x + (Product of Zeros)

P(x) = x² – (1)x + (-6)

P(x) = x² – x – 6

So, one possible quadratic polynomial is x² – x – 6. You can verify that 3 and -2 are indeed its zeros.

  • A quadratic polynomial is of degree 2, in the form ax² + bx + c (a ≠ 0).
  • Zeros of a polynomial are values of ‘x’ that make the polynomial zero.
  • Graphically, zeros are x-intercepts of the parabola y = P(x).
  • Relationship between zeros (α, β) and coefficients (a, b, c): α + β = -b/a, αβ = c/a.
  • Methods to find zeros: Factorization, Quadratic Formula.
  • Discriminant (D = b² – 4ac) determines the nature of roots:
    • D > 0: Two distinct real roots.
    • D = 0: One real root (repeated).
    • D < 0: No real roots.
  • You can form a quadratic polynomial from its zeros using P(x) = k(x – α)(x – β).