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JEE Main 2026 Preparation: Question Papers, Solutions, Mock Tests & Strategy Unacademy » JEE Study Material » Mathematics » Roots and Coefficient of the Quadratic Equation

Roots and Coefficient of the Quadratic Equation

In this article, we will learn about the relation between roots and coefficient of the quadratic equation and the sum of roots of the biquadratic equation.

Table of Content
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The terms from a quadratic equation or polynomial are roots and coefficients. There is a connection between them as well. However, we must first demonstrate the relationship between roots & coefficients for various polynomials. We must understand the meaning of these phrases in order to comprehend the relationship. In the next section, we’ll look at how to derive the relationship between roots and coefficients.

What are the quadratic equation’s roots?

 The roots of a quadratic equation are those values that, when substituted for a variable in a quadratic equation, result in zero. In other words, when the roots of the quadratic equation are substituted for the variable in the quadratic equation, the value equals zero. For instance: If ax2 + bx + c = 0, a real integer α is called the root of the quadratic equation ax2 + bx + c = 0. 

We declare that x = α fulfils the equation ax2 + bx + c = 0 if is a root of ax2 + bx + c = 0. Please notice that the zeros of the polynomial ax2 + bx + c are the roots of the quadratic equation ax2 + bx + c = 0. The 2 zeros or roots of quadratic equation ax2 + bx + c = 0 are α and β.

α = (-b + b2 – 4ac) / 2a and β = (-b – b2 – 4ac) / 2a are the results. 

Please keep in mind that b2 – 4ac is also known as the discriminant. D = b2 – 4ac is the symbol for it. The zeros of the polynomial are the roots of the quadratic equation, as we know. So Polynomial roots are expressed as the sum and product of zeros.

 Let the zeros of a polynomial ax2 + bx + c be α and β

Sum of the zeros (α+β) = {(-b + b2 – 4ac) + (-b – b2 – 4ac)} / 2a}

 (α + β) = – b /a 

Product of the zeros (α.β) = (-b + b2 – 4ac) / 2a (-b – b2 – 4ac) / 2a (-b – b2 – 4ac) / 2a 

(α.β) = c /a

The origins of roots:

When D > 0 (case 1): The roots are real and distinct in this example. 

When D = 0 (case 2): Roots are real and equal in this scenario. 

When D<0 (case 3): the roots are fictitious in this case, thus we say the equation has no actual roots.

What are quadratic equation coefficients? 

Quadratic polynomial coefficients are constant numbers that will never change. 

P(x) = ax2 +bx + c = 0 is an example of a polynomial. The constant terms a, b, and c are also known as the coefficients. Please keep in mind that a ≠ 0.

 In other words, a polynomial is an expression of the type ax2 + bx + c = 0, where a 0 is the coefficient and x2, x is the variable.

Relationship between quadratic equation roots and coefficients:

 The following is the relationship between roots and coefficients: The sum of the roots in a quadratic equation is equal to the negative of the second term’s coefficient divided by the first term’s coefficient. The third term divided by the first term equals the product of roots.

 Let the zeros of a polynomial ax2 + bx + c be α and β.

 Sum of the zeros (α+β) = (-b + b2 – 4ac) / 2a + (-b – b2 – 4ac) / 2a 

(α+β) = – b /a 

Product of the zeros (α.β) = {(-b + b2 – 4ac) / 2a} x {(-b – b2 – 4ac) / 2a}

(α.β) = c /a

We’ll look at the polynomial ax2 + bx + c = 0. Because this polynomial is quadratic, there are only two potential zeros. Which we consider the name and. 

Now, if α and β are the polynomial’s zeros, then (x -α) and (x -β) are the polynomial’s factor ax2 + bx + c = 0. As a result, this polynomial must equal the product of these two variables. 

∴ ax2 + bx + c = (x – 𝛂) (x -𝛃) 

To compare polynomials, multiply any constant term in the right side. 

  • ax2 + bx + c = k (x -α) (x -β)
  • ax2 + bx + c = k. {x2 – (α + β) x + αβ}
  • ax2 + bx + c = kx2 – k (α+ β) x + k(αβ) x 

When the coefficients of like powers of x on both sides are compared, we get 

k = a, – k (α+β) = b, and k(αβ) = c.

Because k Equals a, replace ‘k’ with ‘a.’ 

  •  – a(α+β) = b and a(αβ) = c 
  •  (α+β) = – b/a and (αβ) = c/a 

As a result, the sum of zeros = -(coefficient of x) divided by coefficient of x2. 

Product of zeros = constant term divided by x2 coefficient.

Conclusion:

There are two different real roots when the discriminant is higher than 0. There is just one real root when the discriminant is equal to 0. There are no real roots when the discriminant is smaller than zero, but there are exactly two different imaginary roots. There is only one true root in this scenario. A quadratic equation’s roots are equal to the negation of the second term’s coefficient divided by the leading coefficient. The constant term (the third term) divided by the leading coefficient is the product of the roots of a quadratic equation.

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Frequently Asked Questions

Get answers to the most common queries related to the JEE Examination Preparation.

Find the zeros of the polynomial f(x) = x2 + 7x + 12 and check the relationship between its zeros and coefficients.

 Answer: We have f(x) = x2 + 7x + 12 ...Read full

Find the zeros of the polynomial f(x) = 2x2 + 5x - 12 and check the relationship between its zeros and coefficients.

 Answer: We can deduce that f(x) = 2x2 ...Read full

Find the value of k if the difference between the roots of the equation x2 - 13x + k = 0 is 17.

Answer:  x2 – 13x + k =0...Read full

What's the connection between the root and the coefficient?

Answer: Let the roots of the above equation be α and β. We are sometimes given the relation between the roots of a...Read full

What is the quadratic equation coefficient?

Answer: The quadratic term’s coefficient, a, influences how broad or narrow the graphs are, as well as whether...Read full

 Answer: We have f(x) = x2 + 7x + 12 

                                        = x (x + 4) + 3(x + 4) 

                                        = (x + 4) (x + 3)

  •  f(x) = 0 = (x + 4) (x + 3) = 0

∴ x = -4 and x = -3 

As a result, the zeros of the polynomial f(x) = x2 + 7x + 12 are -4 and -3. 

Product of zeros = (-4). (-3) = 12/1 

Sum of zeros = -4) + (-3) = -7/1 

 Answer: We can deduce that f(x) = 2x2 + 5x – 12

                                             = 2x2 + 8x – 3x – 12 

                                             = 2x (x + 4) – 3(x + 4) 

                                             = (x + 4) (2x – 3) 

  •  f(x) = 0 = (x + 4) (2x – 3) = 0 

∴ x = -4 and x = 3/2  

Sum of the zeros = -4 + 3/2 = -5/2 

Product of the zeros = (-4) x (3/2) = -12/2 

Answer:  x2 – 13x + k =0

 Here, a = 1, b = 13, and c = k 

Let the roots of the equation be α and β.

Then, α+β = -b/a = -(-13)/1 = 13 …… (1)

 also (α –β) = 17…. (2) 

(1) +(2) gives us 2α = 30 => α= 15

As a result, 15 +β= 13 (from (1)) 

  • Β = -2

However, because αβ = c/a = k/1 provides 15 x (-2) = k, we get k = -30.

Answer: Let the roots of the above equation be α and β. We are sometimes given the relation between the roots of a quadratic equation and asked to identify the condition, which is the relation between the coefficients a, b, and c of the quadratic equation.

 Using the formulas α+β = -ba and αβ= ca, this is simple to accomplish.

Answer: The quadratic term’s coefficient, a, influences how broad or narrow the graphs are, as well as whether they curve upward or downward. The ends of the parabola point upward when the quadratic coefficient is positive. The ends of the parabola point downward when the quadratic coefficient is negative.

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