Quadratic Equation

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Complex numbers are the subsets of the set of the real numbers and the set of the imaginary numbers. It is expressed in the form of a+bi. I have chosen this topic as this itself speaks about complex and I have taken this as a challenge as I got always less marks in exams in this topic complex numbers only and I want to improve my weak point. So, I have chosen this as my exploration question as a challenge so that I can have knowledge about that and it will automatically helps me in developing my weakness.

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Also, I am following the IB learner profile and that is risk taker who is enthusiastic to scrutinize new ideas.

In this exploration, I will graphically interpret the complex roots of a quadratic equation by using the real roots of quadratic equation. So, I used different methods to calculate the roots of various quadratic equations and to show graphically, I have also used different graph software's.

Additionally, I have also done the comparison between real and complex roots of quadratic equation.

PROCEDURE

First of all, before talking about complex roots it is important to know about real roots of quadratic equation. The quadratic equation is an equation whose general form is ax 2+bx+c and there are various methods to find the roots of it.

Numerically,
Middle-term splitting
Quadratic formula
Completing the square method

For example: x^2-5x+6=0

Factorization method: x^2-5x+6=0

x^2-2x-3x+6=0

x(x-2)-3(x-2)=0

(x-3)(x-2)=0

x =3 , x =2

By using Quadratic Formula:

x=(-b±?(b^2-4ac))/2a

Here, a=1, b=-5, c=6

Now, x=(-b±?(b^2-4ac))/2a

x=(5±?(?(-5)?^2-4*1*6))/(2*1)

x=(5±?(25-24))/2

x=(5±1)/2

x=(5+1)/2 , x=(5-1)/2

x=6/2 , x=4/2

x=3 , x=2

Completing the square method:

? x?^2-5x+6=0

x^2-5x = -6

x^2-5x +?((-5)/2)?^2= -6+?((-5)/2)?^2

?(x-5/2)?^2 = -6+25/4

?(x-5/2)?^2 = (-24+25)/4

?(x-5/2)?^2= 1/4

x-5/2 =(±1)/2

x = 1/2+5/2 , x= -1/2+5/2

x= 6/2 , x= 4/2

x=3 , x=2

Graphically,

x^2-5x+6=0

This is also one of the ways to find the roots of quadratic equation when it crosses the x-axis as the real root of f(x) = x^2-5x+6 is where the graph y=f(x) cuts the x-axis.

So, in figure 1, the roots are 2 and 3. When there is one solution then the graph just meets the x axis and when there are two solutions, the graph crosses the x axis. In contrast, when the graph does not cross the x-axis, we get imaginary roots for the value of x in the equation and we are interested in complex roots of quadratic equation only.To understand it more clearly, here in figure - 2, there are many other equations including the above quadratic equation.

So, this graph clearly shows 4 equations which cross the x axis at two points, one equation which touch the x axis at a single point and 5 equations which don't touches the x axis. To be more clearly about the complex number plane, the x axis shows the real roots whereas the y axis shows the imaginary roots. Also, the complex roots can be calculated by using a quadratic formula.

Equations Real or Complex Roots

2x^2+15x+15 Real roots

16x^2+32x+16 Real roots

4x^2-5x-6 Real roots

2x^2-13x+14 Real roots

x^2-5x+6 Real roots

3x^2-19x+48 Complex roots

5x^2+5x+5 Complex roots

? x?^2-12x+52 Complex roots

x^2-2x+21 Complex roots

x^2-7x+15 Complex roots

Let's calculate the roots of these equations:

2x^2+15x+15

a = 2, b = 15, c = 15

D = b^2-4ac

D = ?(15)?^2-4*2*15

D = 225-120

D = 105

Now, x=(-b±?(b^2-4ac))/2a

x=(-15±?105)/(2*2)

x=(-15±10.25)/4

x=(-15)/( 4) ± 10.25/4

x= -3.75 ± 2.56

x= ?3.75 + 2.56, x = -3.75 - 2.56

x= ?1.19, x= -6.31

2.) 16x^2+32x+16

a = 16, b = 32, c = 16

D = b^2-4ac

D = ?(32)?^2-4*16*16

D = 1024-1024

D = 0

Now, x=(-b±?(b^2-4ac))/2a

x=(-32±?0)/(2*16)

x=(-32±0)/32

x=(-32)/( 32) ± 0/32

x= -1 ± 0

x= ?1

3.) 4x^2-5x-6

a = 4, b = -5, c = -6

D = b^2-4ac

D = ?(-5)?^2-4*4*-6

D = 25+96

D = 121

Now, x=(-b±?(b^2-4ac))/2a

x=(5±?121)/(2*4)

x=(5±11)/8

x=(5+11)/( 8), x= (5-11)/8

x= 16/( 8), x= (-6)/( 8)

x= 2, x = -0.75

4.) 2x^2-13x+14

a = 2, b = -13, c = 14

D = b^2-4ac

D = ?(-13)?^2-4*2*14

D = 169-112

D = 57

Now, x=(-b±?(b^2-4ac))/2a

x=(13±?57)/(2*2)

x=(13±7.55)/4

x=13/( 4) ± 7.55/4

x= 3.25 ± 1.89

x= 3.25 + 1.89, x = 3.25 - 1.89

x= 5.14, x= 1.36

5.) x^2-5x+6

We have already calculated the roots of this equation earlier.

6.) 3x^2-19x+48

a = 3, b = -19, c = 48

D = b^2-4ac

D = ?(-19)?^2-4*3*48

D = 361-576

D = -215

Now, x=(-b±?(b^2-4ac))/2a

x=(19±?((-215)))/(2*3)

x=(19±14.66i)/6

x=19/6 ± 14.66/6 i

x= 3.17 ± 2.44i

7.) 5x^2+5x+5

a = 5, b = 5, c = 5

D = b^2-4ac

D = ?(5)?^2-4*5*5

D = 25-100

D = -75

Now, x=(-b±?(b^2-4ac))/2a

x=(-5±?((-75)))/(2*5)

x=(-5±8.66i)/10

x=(-5)/( 10) ± 8.66i/10

x= -0.5 ± 0.86i

8.) x^2-12x+52

a =1, b = -12, c = 52

D = b^2-4ac

D = ?(-12)?^2-4*1*52

D = 144-208

D = -64

Now, x=(-b±?(b^2-4ac))/2a

x=(12±?(-64))/(2*1)

x=(12±8i)/2

x=12/2 ± 8/2 i

x= 6 ± 4i

9.) x^2-2x+21

a = 1, b = -2, c = 21

D = b^2-4ac

D = ?(-2)?^2-4*1*21

D = 4-84

D = -80

Now, x=(-b±?(b^2-4ac))/2a

x=(2±?((-80)))/(2*1)

x=(2±8.94i)/2

x=2/( 2) ± 8.94/2 i

x= 1 ± 4.47i

10.) x^2-7x+15

a=1 , b=-7 , c=15

Now, ?D = b^2-4ac

(-7)^2-4*1*15

49 60

-11

To calculate roots, x=(-b±?D)/2a

x=(7±?((-11)))/2

x= 3.5 ± 5.5i

The i is an imaginary number and it is equal to?(-1) From this, we can conclude that when the discriminant (D) > 0, we get two real and distinct roots. & when D = 0, we get two real and equal roots. & when D < 0, we get two imaginary roots. As we are interested in graphical interpretation of complex roots of quadratic equation, so to configure we will start with the general form of quadratic equation ax^2+bx+c by completing the square method and by using the real roots of quadratic equation.

y= ax^2+bx+c

a(x^2+b/a x)+c

?a((x+b/2a)?^2-?(b/2a)?^2)+c

?a(x+b/2a)?^2-b^2/4a+c

To reflect the quadratic:

?y=-a(x+b/2a)?^2-b^2/4a+c

To compare it with real quadratic, we have to simplify and then put the coefficients of the equation to a quadratic formula.

y= -a(x^2+(b/2a)^2+2*x*b/2a)-b^2/4a+c

-a(x^2+b^2/?4a?^2 +2bx/2a)-b^2/4a+c

-a(x^2+b^2/?4a?^2 +bx/a)-b^2/4a+c

-ax^2-?ab?^2/?4a?^2 -abx/a-b^2/4a+c

-ax^2-b^2/4a-bx-b^2/4a+c

-ax^2-bx-?2b?^2/4a+c

Now, x=(b±?(b^2-4(-a)(-?2b?^2/4a+c)))/(-2a)

x=(b±?(b^2-4a((?2b?^2+4ac)/4a)))/(-2a)

x=(b±?(?-b?^2+4ac))/(-2a)

x=(-b±i?(b^2-4ac))/2a

Here, we can see the very slight difference between real and complex roots of quadratic equations. Only i (the imaginary number) is there in this quadratic formula. Graphically, a reflection to the graph of real roots will produce another new equation that would have complex roots.

Let's have example,

y=5x^2+3x+6

To be further precisely, we will take another equations:

y=3x^2+5x+3

In addition, y=2x^2+2x+1

In contrast, we can also find the real roots through the reflection of complex roots of quadratic equations by using completing the square method only. So, here is the equation:

y=-2x^2+x-1

CONCLUSION:

We conclude that we found x units in real direction and i units in both positive and negative directions. When the discriminant (b^2-4ac) is negative, we get two complex solutions and we can graphically interpret the complex roots of quadratic equation through the reflection of real roots of quadratic equation. So, both are very important and there are abundant real and complex roots. It deals with our daily life situations as well.