Quadratic Equations: Are there Unreal Roots?

Roots of a Quadratic Equation:

Quadratic equations in standard form are represented as, \(ax^2~+bx~+~c\) = 0, where a ≠ 0 and a,b,c are real constants. A real number x which satisfies the equation \(ax^2~+~bx~+~c\) = 0,is known as its root or solution.

The solutions or the roots can be found either by factorization method or by using the formula:

x = \( \frac {-b~±~\sqrt{b^2~-~4ac}}{2a} \)

⇒ x = \( \frac -{b}{2a}~±~\frac{\sqrt{b^2~-~4ac}}{2a} \)

The term \( b^2~-~4ac\) is known as Discriminant of the quadratic equation \(ax^2~+~bx~+~c\)= 0 and is denoted by Dor Δ.

The nature of the roots of quadratic equation can be figured out using the discriminant.

If D = 0, then the equation has two equal roots which is equal to \(\frac{–b}{2a}\).

If D > 0, then the equation has two distinct roots, given by \( \frac {-b~+~\sqrt{b^2~-~4ac}}{2a} \)and \( \frac {-b~-~\sqrt{b^2~-~4ac}}{2a} \). The roots can either be rational or irrational.

Now, consider the situation where discriminant, \(b^2~-~4ac\) < 0.For such a case, the root will become:

x = \(- \frac{b}{2a} ~±~\frac{\sqrt{Negative ~number}}{2a} \)

As the square root of any negative number is not defined for real numbers we need another system of numbers which would help us figure out the root of a negative number.

In mathematics, natural numbers were the first one to come into existence. Then we realized the need to represent nothing, hence zero was discovered. Then followed negative numbers, fractions. Until then it was thought that rational numbers covered all the numbers possible. Then the Pythagorean’s, who were obsessed with triangles were once trying to find the hypotenuse of a right angled isosceles triangle which had base and perpendicular sides equal to 1 unit. No rational number could define it as measure of hypotenuse would be \( \sqrt{2} \). Hence irrational numbers were discovered.

Similarly, while solving quadratics,need for the kind of numbers which could define root of negative numbers was realized. Therefore imaginary numbers were discovered which could define the root of a negative number. These numbers are represented on an imaginary number line.

Any numberz which can be represented in the form of z = x + iy where x and y are both real numbers, x is the real part and y is the imaginary part of complex number z and i is an imaginary unit satisfying the equation \(i^2~=~-1\).


Figure 1 Complex Number representation

So, the concept of complex numbers came into existence.

Consider the quadratic equation,\(x^2~+~1\)= 0, the discriminant \(b^2~-~4ac\) is less than zero.

\(x^2\)= -1

x = \( \sqrt{-1} \)

In the system of complex numbers, we denote \( \sqrt{-1} \) as i(read as iota). i is an imaginary number Therefore, the root of the equation is i.

To find the roots the equation \(x^2~+~3x~+~4\)= 0, let’s calculate its determinant. Here

D = \(3^2~-4~×~1~×~4\) = -7

The roots are given by: x = \( – \frac {b}{2a}~ ± ~ \frac{\sqrt{D}}{2a} \)

Now, \( \sqrt{D} \) = \( \sqrt{-7} \)

This can be written as: \( \sqrt{D} \) = \( \sqrt{-1} \) × \( \sqrt{-7} \)

where \( \sqrt{-1} \)) = i

That is where we need to use complex numbers in algebra.

Let us now have a look whether we can actually plot a graph for complex roots. Consider the equation:

(y-3)= \((x-1)^2 \) —(1)

To plot it let us compare this to (y – k)= a \((x~-~h)^2 \)

By comparison: k = 3,a = 1 and h = 1

As a is positive, the graph will be an upward facing parabola with vertex at (h,k) i.e., (1,3) as shown in given figure.


On expanding equation 1, we get;

y – 3 = \(x^2~+~1~-~2x\)

y = \(x^2~-~2x~+~4\)

Thus y is a function in x. The roots of any equation are defined as the all the possible values which will satisfy the given equation.

The roots of quadratic function f(x) = \(ax^2~+~bx~+~c\) , corresponds to solution of the quadratic equation \(ax^2~+~bx~+~c\) = 0, as it denotes the value of x for which f(x)=0.

Now, to find D the equation can be written as:

\(x^2~-~2x~+~4\)= 0

D = \((-2)^2~-~4(1)(4)\)= -12

\( \sqrt{D} \) = i\( \sqrt{12}\)

Now the roots of this equation will come out to be:

x = \( – \frac{b}{2a} ~±~ \frac{\sqrt{D}}{2a}\)

⇒ x = \(\frac{2 ~±~ i\sqrt{12}}{2} \)

Because the roots are complex, we cannot see it intersecting any axes as in case of real roots the parabola intersects the x-axis at two points which are its roots.

If we take the negative of R.H.S of equation 1, we will get a graph facing downwards and intersecting the real axis at two distinct points.

Now we have: (y – 3)=(-1)\((x~-~1)^2\)


Figure 2 Graph of quadratic Equations

The discriminant for equation \(-x^2~+2x~+~2\) = 0 is 12. As D > 0, therefore it has two real and distinct roots which are given as:

x = \( -\frac{b}{2a} ~±~ \frac{\sqrt{D}}{2a}\)

⇒ x = \(\frac{2 ~±~ i\sqrt{12}}{2} \)<


Figure 3 Graph of quadratic equations

Thus nature of roots of quadratic equations can be determined using discriminant D.

Practise This Question

A block of mass 'm' is hanging to a string in an elevator. The elevator is moving up with an acceleration 'a'. What will be the tension in the string?