SOLVING QUADRATICS REAL AND IMAGINARY SOLUTIONS

To solve quadratic equation, we use the following methods.

(1)  Factoring

(2)  Quadratic formula

(3)  Completing the square method

Without solving the quadratic equation completely, we can decide whether it has real and not equal solution, real and equal solution or imaginary solutions.

Using the 

Δ = b2 - 4ac

Nature of roots

b2 - 4ac > 0

b2 - 4ac = 0

b2 - 4ac < 0

Roots are real and unequal

Roots are real and equal

Roots are imaginary

Find roots of the quadratic equations given below.

Example 1 :

x2-6x+12  =  0

Solution :

x2-6x+12  =  0

a  =  1, b  =  -6 and c  =  12

Solving the quadratic equation using the formula

x  =  -b±√(b2-4ac)/2a

we can solve the equation.

b2-4ac  =  6- 4 ⋅ 1 ⋅ 12

=  36 - 48

=  -12 < 0

Since b2-4ac < 0, the quadratic equation will have imaginary roots.

b2-4ac  =  √-12  ==>  2i√3

x  =  (6 ± 2i√3)/2

x  =  (3 ± i√3)

So, the roots are 3 + i√3 and 3 - i√3.

Example 1 :

14-3x2  =  2x

Solution :

14-3x2  =  2x

Multiply by negative.

3x2 - 14  =  -2x

3x2+2x-14  =  0

a  =  3, b  =  2 and c  =  -14

Solving the quadratic equation using the formula

x  =  -b±√(b2-4ac)/2a

we can solve the equation.

b2-4ac  =  2- 4 ⋅ 3 ⋅ (-14)

=  4 + 168

=  172 > 0

Since b2-4ac > 0, the quadratic equation will have real roots.

b2-4ac  =  √172  ==>  2√43

x  =  (2 ± 2√43)/2

x  =  (1 ± √43)

So, the roots are (1+√43)/2 and (1-√43)/2.+

Find a possible pair of integer values for a and c so that the quadratic equation has the given solution(s). 

Example 3 :

ax2+4x+c  =  0 has two imaginary solution.

Solution :

Since the quadratic equation has imaginary roots.

b2-4ac < 0

a  =  a, b  =  4 and c  =  c

42-4ac < 0

16-4ac < 0

To  get the imaginary  solutions, 4ac should be > 16

So, product of a and c should be > 4

One of the possible value If a  =  2 and b  =  3

So, the required equation is 2x2+4x+3  =  0

Example 4 :

 ax2 − 8x + c = 0 two real solutions

Solution :

Since the quadratic equation has imaginary roots.

b2-4ac > 0

a  =  a, b  =  -8 and c  =  c

(-8)2-4ac > 0

64-4ac > 0

To  get the real solutions, 4ac should be > 64

So, product of a and c should be < 16

One of the possible value If a  =  2 and b  =  3

So, the required equation is 2x2 − 8x + 3 = 0

Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. SAT Math Resources (Videos, Concepts, Worksheets and More)

    Oct 01, 24 12:04 PM

    SAT Math Resources (Videos, Concepts, Worksheets and More)

    Read More

  2. Digital SAT Math Problems and Solutions (Part - 47)

    Oct 01, 24 11:49 AM

    Digital SAT Math Problems and Solutions (Part - 47)

    Read More

  3. Digital SAT Math Problems and Solutions (Part - 46)

    Sep 30, 24 11:38 AM

    digitalsatmath41.png
    Digital SAT Math Problems and Solutions (Part - 46)

    Read More