Nature of Roots of a Quadratic Equation: Formula, Examples

Let us learn about the Nature of the Roots of a Quadratic Equation. A quadratic equation is an equation of degree 22. When a polynomial is equated to zero, we get an equation known as a polynomial equation. If a quadratic polynomial is equated to zero, we can call it a quadratic equation. 

A quadratic equation represents a parabolic graph with two roots. Such equations arise in many real-life situations such as athletics(shot-put game), measuring area, calculating speed, etc. This article will explain the nature of the roots formula and understand the nature of their zeros or roots.

Definition of Quadratic Equations

An equation of second-degree polynomial in one variable, such as \(x\) usually equated to zero, is a quadratic equation. The coefficient of \(x^2\) must not be zero in a quadratic equation. If \(p(x)\) is a quadratic polynomial, then \(p(x)=0\) is called a quadratic equation.

For example, \({x^2} + 2x + 2 = 0\), \(9{x^2} + 6x + 1 = 0\), \({x^2} – 2x + 4 = 0,\) etc are quadratic equations.

General Form of a Quadratic Equation

The general form of a quadratic equation is given by \(a{x^2} + bx + c = 0,\) where \(a, b, c\) are real numbers, \(a \ne 0\) and \(a\) is the coefficient of \(x^2,\) \(b\) is the coefficient of \(x,\) and \(c\) is a constant.

Roots of Quadratic Equations

The values of the variable \(x\) that satisfy the equation in one variable are called the roots of the equation. These roots may be real or complex. If a quadratic polynomial is equated to zero, it becomes a quadratic equation. The values of \(x\) satisfying the equation are known as the roots of the quadratic equation.

In general, a real number \(α\) is called a root of the quadratic equation \(a{x^2} + bx + c = 0,\) \(a \ne 0.\) If \(a{\alpha ^2} + b\alpha + c = 0,\) we can say that \(x=α\) is a solution of the quadratic equation. Note that the zeroes of the quadratic polynomial \(a{x^2} + bx + c\) and the roots of the quadratic equation \(a{x^2} + bx + c = 0\) are the same.

The roots of the quadratic equation \(a{x^2} + bx + c = 0\) are given by
\(x = \frac{{ – b \pm \sqrt {{b^2} – 4ac} }}{ {2a}}\)
This is the quadratic formula for finding the roots of a quadratic equation.

Discriminant

Consider a quadratic equation \(a{x^2} + bx + c = 0,\) where \(a\) is the coefficient of \(x^2,\) \(b\) is the coefficient of \(x\), and \(c\) is the constant. The value of \((b^2 – 4ac )\) in the quadratic equation \(a{x^2} + bx + c = 0,\) \(a \ne 0\) is known as the discriminant of a quadratic equation. The discriminant of a quadratic equation determines the nature of roots. Discriminant can be represented by \(D.\)

Nature of the Roots of Quadratic Equation Notes

The value of the discriminant, \(D = {b^2} – 4ac\) determines the nature of the roots of the quadratic equation. If \(a, b, c ∈ R,\) then the roots of the quadratic equation can be real or imaginary based on the following criteria:

The roots are real when \(b^2 – 4ac≥0\) and the roots are imaginary when \(b^2 – 4ac<0.\) We can classify the real roots in two parts, such as rational roots and irrational roots. Let us know about them in brief.

Rational Root

If a quadratic equation is given by \(a{x^2} + bx + c = 0,\) where \(a,b,c\) are rational numbers and if \(b^2 – 4ac>0,\) i.e., \(D>0\) and a perfect square, then the roots are rational.

Irrational Roots

If a quadratic equation is given by \(a{x^2} + bx + c = 0,\) where a,b,c are rational numbers and if \(b^2 – 4ac>0,\) i.e., \(D>0\) and not a perfect square, the roots are irrational.

We can classify the roots of the quadratic equations into three types using the concept of the discriminant.
1. Two distinct real roots
2. Two equal real roots
3. No real roots

Let us discuss the nature of roots in detail one by one.

Two Distinct Real Roots

In a quadratic equation \(a{x^2} + bx + c = 0,\) there will be two roots, either they can be equal or unequal, real or unreal or imaginary.

We can get two distinct real roots if \(D = {b^2} – 4ac > 0.\)

We can represent this graphically, as shown below. In the graphical representation, we can see that the graph of the quadratic equation cuts the \(x\)- axis at two distinct points. These two distinct points are known as zeros or roots.

For example, consider the quadratic equation \({x^2} – 7x + 12 = 0.\)
Here, \(a=1\), \(b=-7\) & \(c=12\)
Discriminant \(D = {b^2} – 4ac = {( – 7)^2} – 4 \times 1 \times 12 = 1\)

Since the discriminant is greater than zero \({x^2} – 7x + 12 = 0\) has two distinct real roots.
We can find the roots using the quadratic formula.
\(x = \frac{{ – ( – 7) \pm 1}}{{2 \times 1}} = \frac{{7 \pm 1}}{2}\)
\( = \frac{{7 + 1}}{2},\frac{{7 – 1}}{2}\)
\( = \frac{8}{2},\frac{6}{2}\)
\(= 4, 3\)

The graph of this quadratic equation cuts the \(x\)-axis at two distinct points.

Two Equal Real Roots

In a quadratic equation \(a{x^2} + bx + c = 0,\) we get two equal real roots if \(D = {b^2} – 4ac = 0.\) In the graphical representation, we can see that the graph of the quadratic equation having equal roots touches the x-axis at only one point. This point is taken as the value of \(x.\)

On the other hand, we can say \(x\) has two equal solutions.

For example, Consider \({x^2} – 2x + 1 = 0.\) The discriminant \(D = {b^2} – 4ac = {( – 2)^2} – 4 \times 1 \times 1 = 0\)
Since the discriminant is \(0\), \({x^2} – 2x + 1 = 0\) has two equal roots.
We can find the roots using the quadratic formula.
\(x = \frac{{ – ( – 2) \pm 0}}{{2 \times 1}} = \frac{2}{2} = 1\)

The graph of this quadratic equation touches the \(x\)-axis at only one point.

No Real Roots

In a quadratic equation \(a{x^2} + bx + c = 0\), if \(D = {b^2} – 4ac < 0\) we will not get any real roots. The roots are known as complex roots or imaginary roots. In the graphical representation, we can see that the graph of the quadratic equation having no real roots does not touch or cut the \(x\)-axis at any point.

For example, \(3{x^2} + x + 4 = 0,\) has two complex roots as \({b^2} – 4ac = {(1)^2} – 4 \times 3 \times 4 = – 47\) that is less than zero.

Let us understand the concept by solving some nature of roots of a quadratic equation practices problem.

Solved Examples

Q.1. Find the discriminant of the quadratic equation \(2 {x^2} – 4x + 3 = 0\) and hence find the nature of its roots.
Ans: The given equation is of the form \(a {x^2} + bx + c = 0.\)
From the given quadratic equation \(a = 2\), \(b = – 4\) and \(c = 3.\)
The discriminant \({b^2} – 4ac = {( – 4)^2} – (4 \times 2 \times 3) = 16 – 24 = – 8 < 0\)
Therefore, there are no real roots exist for the given quadratic equation.

Q.2. Using the quadratic formula method, find the roots of the quadratic equation \(2{x^2} – 8x – 24 = 0\)
Ans: From the given quadratic equation \(a = 2\), \(b = – 8\), \(c = – 24\)
Quadratic equation formula is given by \(x = \frac{{ – b \pm \sqrt {{b^2} – 4ac} }}{{2a}}\)
\(x = \frac{{ – ( – 8) \pm \sqrt {{{( – 8)}^2} – 4 \times 2 \times ( – 24)} }}{{2 \times 2}} = \frac{{8 \pm \sqrt {64 + 192} }}{4}\)
\(x = \frac{{8 \pm \sqrt {256} }}{4} = \frac{{8 \pm 16}}{4} = \frac{{8 + 16}}{4},\frac{{8 – 16}}{4} = \frac{{24}}{4},\frac{{ – 8}}{4}\)
\( \Rightarrow x = 6, x = – 2\)
Hence, the roots of the given quadratic equation are \(6\) & \(- 2.\)

Q.3. Find the discriminant of the quadratic equation \({x^2} – 4x + 4 = 0\) and hence find the nature of its roots.
Ans: Given, \({x^2} – 4x + 4 = 0\)
The standard form of a quadratic equation is \(a{x^2} + bx + c = 0.\)
Now, comparing the given equation with the standard form we get,
From the given quadratic equation \(a = 1\), \(b = – 4\) and \(c = 4.\)
The discriminant \({b^2} – 4ac = {( – 4)^2} – (4 \times 1 \times 4) = 16 – 16 = 0.\)
Therefore, the equation has two equal real roots.

Q.4. Find the discriminant of the quadratic equation \(2{x^2} + 8x + 3 = 0\) and hence find the nature of its roots.
Ans: The given equation is of the form \(a{x^2} + bx + c = 0.\)
From the given quadratic equation \(a = 2\), \(b = 8\) and \(c = 3\)
The discriminant \({b^2} – 4ac = {8^2} – (4 \times 2 \times 3) = 64 – 24 = 40 > 0\)
Therefore, the given quadratic equation has two distinct real roots.

Q.5. Find the roots of the quadratic equation by using the formula method \({x^2} + 3x – 10 = 0.\)
Ans: From the given quadratic equation \(a = 1\), \(b = 3\), \(c = {- 10}\)
Quadratic equation formula is given by \(x = \frac{{ – b \pm \sqrt {{b^2} – 4ac} }}{{2a}}\)
\(x = \frac{{ – (3) \pm \sqrt {{{(3)}^2} – 4 \times 1 \times ( – 10)} }}{{2 \times 1}} = \frac{{ – 3 \pm \sqrt {9 + 40} }}{2}\)
\(x = \frac{{ – 3 \pm \sqrt {49} }}{2} = \frac{{ – 3 \pm 7}}{2} = \frac{{ – 3 + 7}}{2},\frac{{ – 3 – 7}}{2} = \frac{4}{2},\frac{{ – 10}}{2}\)
\( \Rightarrow x = 2,\,x = – 5\)
Hence, the roots of the given quadratic equation are \(2\) & \(- 5.\)

The nature of roots of quadratic equation facts discussed in the above examples will help apply the concept in questions.

Summary

In this article, we discussed the quadratic equation in the variable \(x\), which is an equation of the form \(a{x^2} + bx + c = 0\), where \(a,b,c\) are real numbers, \(a ≠ 0.\) Also, we discussed the nature of the roots of the quadratic equations and how the discriminant helps to find the nature of the roots of the quadratic equation.

Frequently Asked Questions (FAQs)

You can take the nature of the roots of a quadratic equation notes from the below questions to revise the concept quickly.

Q.1. How do you find the nature of the roots of a quadratic equation?
Ans: Since \(\left({{b^2} – 4ac} \right)\) determines whether the quadratic equation \(a{x^2} + bx + c = 0\) has real roots or not, \(\left({{b^2} – 4ac} \right)\) is called the discriminant of this quadratic equation.
So, a quadratic equation \(a{x^2} + bx + c = 0\) has
1. Two distinct real roots, if \({b^2} – 4ac > 0\)
2. Two equal real roots, if \({b^2} – 4ac = 0\)
3. No real roots, if \({b^2} – 4ac < 0\)

Q.2. Explain the nature of the roots of the quadratic Equations with examples?
Ans: Let us take some examples and explain the nature of the roots of the quadratic equations. Consider, \({x^2} – 4x + 1 = 0.\)
The discriminant \(D = {b^2} – 4ac = {( – 4)^2} – 4 \times 1 \times 1 \Rightarrow 16 – 4 = 12 > 0\)
So, the roots of the equation are real and distinct as \(D > 0.\)
Consider, \({x^2} + 6x + 9 = 0\)
The discriminant \({b^2} – 4ac = {(6)^2} – (4 \times 1 \times 9) = 36 – 36 = 0\)
So, the roots of the equation are real and equal as \(D = 0.\)
Consider, \(2{x^2} + x + 4 = 0\), has two complex roots as \(D = {b^2} – 4ac \Rightarrow {(1)^2} – 4 \times 2 \times 4 = – 31\) that is less than zero.

Q.3. What is the nature of a root?
Ans: The values of the variable such as \(x\) that satisfy the equation in one variable are called the roots of the equation. These roots may be real or complex. We can classify the zeros or roots of the quadratic equations into three types concerning their nature, whether they are unequal, equal real or imaginary. To determine the nature of the roots of any quadratic equation, we use discriminant.

Q.4. What are the five real-life examples of a quadratic equation?
Ans: Five real-life examples where quadratic equations can be used are
(i) Throwing a ball
(ii) A parabolic mirror
(iii) Shooting a cannon
(iv) Diving from a platform
(v) Hitting a golf ball
In all these instances, we can apply the concept of quadratic equations.

Q.5. How can you tell if it is a quadratic equation?
Ans: An equation is a quadratic equation in the variable \(x\) if it is of the form \(a{x^2} + bx + c = 0\), where \(a, b, c\) are real numbers, \( a ≠ 0.\)

Q.6. What is the standard form of the quadratic equation?
Ans: The form \(a{x^2} + bx + c = 0,\) \( a ≠ 0\) is called the standard form of a quadratic equation.

Q.7. What is a discriminant in a quadratic equation?
Ans: The term \(\left({{b^2} – 4ac} \right)\) in the quadratic formula is known as the discriminant of a quadratic equation \(a{x^2} + bx + c = 0,\) \( a ≠ 0.\) The discriminant of a quadratic equation shows the nature of roots.

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