Can you solve all the quadratic equations? Can quadratic equations have more than one solution? Does any quadratic equation exist that has no real solution? The value of the variable for which the equation agrees is known as the solution or roots of the quadratic equation.

The number of roots of a quadratic equation is equivalent to its degree. Thus, the quadratic equation has two roots.

Some Methods of Finding Roots of a Quadratic Equation are:

Factorization method

Quadratic Formula method

Completing the square method.

The concept of the nature of quadratic equations is quite interesting. Let us learn the meaning of nature of the roots of the quadratic equation, nature of the roots of the quadratic equation formula, graphical representation of quadratic equation, nature of the roots of the quadratic equation example, etc.

Quadratic equations are considered as a polynomial equation of degree 2 in one variable of type f(x) = ax2 +bx +c =0, where a,b,and c are real numbers and a ≠ 0. It is a general expression of quadratic equations, where a is considered as a leading coefficient and c is considered as the absolute term of f(x). The values of x satisfying the quadratic equation are known as the roots of the quadratic equation.

The quadratic equation will always have two roots. The nature of the roots can be either real or imaginary.

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A polynomial in the form of ax2 +bx +c =0, where a,b and c are real numbers and a ≠ 0 is known as a quadratic polynomial. A quadratic polynomial becomes a quadratic equation when it is equated to 0.

ax2 +bx +c =0,

Examples: 4x2 +x+7, -2x2 +6x+6, 2x2+x=0

The value of x for which a quadratic equation agrees is known as roots of the quadratic equation.

If α is a roots of the quadratic equation the expression ax2 +bx +c =0, then, aα 2 + bα +c =0

A quadratic equation can have two different roots, two similar roots or real roots may not exist.

Every quadratic equation with real roots can be factorized. The graph of the equation intersects at the x-axis at the root of an equation.The x-axis signifies the real line in the Cartesian plane. It implies that the equation with unreal roots won't intersect at the x-axis and it can not even be written in a factored form.

The roots of the quadratic equation are the points where the graph of the quadratic polynomial touches the x- axis.

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In the graph given above, -2 and 2 are the roots of the quadratic equation x2-4=0

Important Points

If the graph of the quadratic polynomial equation touches the x-axis at two different points, then it implies that the equations have real and distinct roots.

If the graph of the quadratic polynomial equation touches the x-axis, then it implies that it has real and equal roots.

If the graph of the quadratic polynomial equation does not touch the x-axis, then it implies that it does not retain any real roots.

Now, we will study the concept of the nature of the roots of a quadratic equation by considering the following equation:

ax2+bx + c= 2

The roots are given through the quadratic formula for the equation given above:

x = \[\frac{-b ± \sqrt{b^{2}-4ac}}{2a}\].

Let us take a real number p > 0. Now, as we know\[\sqrt{p}\] is defined and a positive number.

Is \[\sqrt{-k}\] a real number ? The answer is no. For example : If we have \[\sqrt{225}\], we can write it as \[\sqrt{15\times15}\]which is equivalent to 15.

If we have \[\sqrt{-225}\]. We can never write this number as a product of two similar quantities. It is because it has negative signs which can only result from the product of two quantities having opposite signs.

The part which is under the square root in the expression of root such as b2 -4ac is known as a discriminant of the quadratic equation.This is the value that discriminates against other quadratic equations having distinct roots. It is denoted by D.So,

D =b2 - 4ac

On the basis of D, the roots can be expressed as:

x= - b ±\[\sqrt{D}\]/ 2a

Now, consider D as a real number since x,y,z are real numbers. On the basis of the value of x,y and x,z, the value of Z can be either positive, negative or zero. Let us examine all its possibilities and how it affects the roots of quadratic equations.

X = -b +\[\sqrt{D}\]/ 2a or -b -\[\sqrt{D}\]/ 2a

D > 0 -The equation will have two real and distinct roots when D is positive. It implies that the graph of the quadratic equation will intersect the x- axis at two distinct points.

D = 0,-The equation will have two real and two equal roots when D= 0. It implies that the graph of the equation will intersect the x-axis exactly at one single point. The roots can be easily examined for the equation y by substituting D= 0. The roots are:

x = -b/2a

D < 0,-The equation will have no real roots when D is negative. It implies that the graph of the quadratic equation will not intersect x-axis.

The roots of the quadratic equation are derived by the following quadratic formula.

x = \[\frac{-b ± \sqrt{b^{2}-4ac}}{2a}\]

Below given, the nature of the roots of the quadratic equation example will help you to understand the concept thoroughly:

Example -1: x2 + 5x + 6

Solution:

D = b2 - 4ac

D = 52 – 4 x 1 x 6 = 25 -24 = 1

D = Since D > 0, the equation will have two real roots and distinct roots. The roots are:

x= - b + √ d / 2a or -b - √ d / 2a]

x= - 5 + √1/ 2 x 1 or - 5 -√1 / 2 x 1

x= -2 or -3

1. Discuss the nature of the roots of a quadratic equation 2x2 – 8x +3 = 0

Solutions: Here, the coefficients are all rational. The discriminant D for a given equation will be

D = b2 – 4ac = (-8)2- 4*2*3

=64-24

= 40 > 0

We can see, the discriminant of the given quadratic equation is positive but not a perfect square.

Hence, the roots of a quadratic equation are real, unequal and irrational.

2. Determine the value of k for which the quadratic expression (x-a) (x-10) +1 =0 has integral roots.

Solution : The given quadratic equation can be rewritten as

x2 – (10 + k) x +1 + 10k = 0

b2 – 4ac = 100 + k2 + 20k – 40k = k2 -100k + 96 = (k - 10)2 - 4

The quadratic equation will have an integral root if the value of discriminant > 0, D is a perfect square, a=1 and b and c are considered as integers.

i.e. (k-10)2 - D =4

Since, the discriminant is a perfect square. Hence, the difference of the two perfect square in R.H.S will be 4 only when D=0 and (k-10)2 =4

→ k-10 = ± 2, Hence, the values of k = 8 and 112.

Sudhara, the Sanskrit philosopher, introduced the formula to find the two roots of a quadratic equation.

The Babylonians introduced a technique known as completing the square to resolve common issues with areas by 400 BC.

The first reliable mathematical try to introduce quadratic formula was done by Pythagoras in 500 BC.

1. Calculate the value (s) of p for which the quadratic equation 2x2 + px + 8 =0 has similar roots

p= ± 64

p= ±18

p= ±4

p= ± 16

2. If the roots 32 + kx + 12=0 are equal , then k =

±12

12

-12

0

3. The roots of the quadratic equation x2 - 6x + 10 = 0 are

Equal

Imaginary

Irrational

real

FAQ (Frequently Asked Questions)

1. Explain the Term Discriminant in Quadratic Equations.

The discriminant is said to be the part of the quadratic formula under the square root.

x = [-b ± √(b²-4ac)] / 2a.

The discriminant can be zero, positive or negative and it states the number of solutions can be given to the quadratic equation.

A positive discriminant states that the quadratic has two different real numbers

A zero discriminant states that quadratic has repeated real number solutions.

A negative discriminant indicates that neither of these solutions is real numbers.

Discriminant example

We need the find out the number of solutions that can be given to the quadratic equation 6x +10x -1

Solution: For the above equation we can see,

a= 6, b= 10, c= -1

Substituting these values in discriminant we get,

b^{2}-4ac

= 10^{2} -4 (6)(-1)

=100 + 24

=124

This is a positive number which implies that it has two solutions.

2. How to Calculate the Roots of a Quadratic Equation?

Roots of a quadratic equation can be calculated by factoring the polynomial and equating it to zero.

Consider the quadratic equation which is in the form of ax^{2} + bx +c =0

Here, -5x is split into two parts -4x-x, as the multiplication of these parts result in 4x^{2} which is equivalent to a *c = 2x^{2} * 2=4x^{2}

Thus, the equation becomes:

2x^{2}-5x +2 =0 =2x^{2} – 4x –x +2

= 2x (x-2) -1 (x-2)

= (2x-1) (x-2)

Therefore, 2x^{2}-5x +2 = 0 is similar as (2x-1) (x-2) = 0

The value of x for which 2x^{2}-5x +2 =0 is similar as the value of x for which (2x-1) (x-2) = 0

If (2x-1) (x-2) = 0, then either (2x-1) = 0 or (x-2) =0

2x-1=0 gives 2x=1 or x=1/2

x-2=0 gives x = 2

Therefore, ½ and 2 are the roots of the equation 2x^{2}-5x +2