Roots of a Quadratic Equation:

Quadratic equations in standard form are represented as, \(ax^2~+bx~+~c\)

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

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

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

The term \( b^2~-~4ac\)

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 \(\frac{â€“b}{2a}\)

If D > 0, then the equation has two distinct roots, given by \( \frac {-b~+~\sqrt{b^2~-~4ac}}{2a} \)

Now, consider the situation where discriminant, \(b^2~-~4ac\)

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

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 \( \sqrt{2} \)

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,\(x^2~+~1\)

\(x^2\)

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

In the system of complex numbers, we denote \( \sqrt{-1} \)

To find the roots the equation \(x^2~+~3x~+~4\)

D = \(3^2~-4~Ã—~1~Ã—~4\)

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

Now, \( \sqrt{D} \)

This can be written as: \( \sqrt{D} \)

where \( \sqrt{-1} \)

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)= \((x-1)^2 \)

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

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

As aÂ 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) = \(ax^2~+~bx~+~c\)

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

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

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

â‡’ \( \sqrt{D} \)

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

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

â‡’ x = \(\frac{2 ~Â±~ i\sqrt{12}}{2} \)

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 \(-x^2~+2x~+~2\)

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

â‡’ x = \(\frac{2 ~Â±~ i\sqrt{12}}{2} \)

Figure 3 Graph of quadratic equations

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