What are imaginary roots? How to find imaginary roots of an equation? What are the difference methods to find imaginary roots?
Find the roots of given equation..
2x² + x + 4 = 0.
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Answers
x^2+1=0. Solving for x yields:
x^2 = -1,
x = sqrt(-1) = i.
hmm, thats not a real number that I can find on the real axis where I am used to finding real roots to equations like this. That’s because its not a real root. Instead, its an imaginary root. It can be found on the plane of Complex numbers. Polynomials can have real roots only, real and imaginary roots too, or only imaginary roots.
This one only has imaginary roots, 2 of them, i and -i.
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Solve Quadratic Equation using the Quadratic Formula
3.3 Solving 2x2+x+4 = 0 by the Quadratic Formula .
According to the Quadratic Formula, x , the solution for Ax2+Bx+C = 0 , where A, B and C are numbers, often called coefficients, is given by :
- B ± √ B2-4AC
x = ————————
2A
In our case, A = 2
B = 1
C = 4
Accordingly, B2 - 4AC =
1 - 32 =
-31
Applying the quadratic formula :
-1 ± √ -31
x = ——————
4
In the set of real numbers, negative numbers do not have square roots. A new set of numbers, called complex, was invented so that negative numbers would have a square root. These numbers are written (a+b*i)
Both i and -i are the square roots of minus 1
Accordingly,√ -31 =
√ 31 • (-1) =
√ 31 • √ -1 =
± √ 31 • i
√ 31 , rounded to 4 decimal digits, is 5.5678
So now we are looking at:
x = ( -1 ± 5.568 i ) / 4
Two imaginary solutions :
x =(-1+√-31)/4=(-1+i√ 31 )/4= -0.2500+1.3919i or:
x =(-1-√-31)/4=(-1-i√ 31 )/4= -0.2500-1.3919i
Two solutions were found :
x =(-1-√-31)/4=(-1-i√ 31 )/4= -0.2500-1.3919i x =(-1+√-31)/4=(-1+i√ 31 )/4= -0.2500+1.3919i
Step-by-step explanation:
x^2+1=0. Solving for x yields:
x^2 = -1,
x = sqrt(-1) = i.
hmm, thats not a real number that I can find on the real axis where I am used to finding real roots to equations like this. That’s because its not a real root. Instead, its an imaginary root. It can be found on the plane of Complex numbers. Polynomials can have real roots only, real and imaginary roots too, or only imaginary roots.
This one only has imaginary roots, 2 of them, i and -i.
.........
......
.....
.....
Solve Quadratic Equation using the Quadratic Formula
3.3 Solving 2x2+x+4 = 0 by the Quadratic Formula .
According to the Quadratic Formula, x , the solution for Ax2+Bx+C = 0 , where A, B and C are numbers, often called coefficients, is given by :
- B ± √ B2-4AC
x = ————————
2A
In our case, A = 2
B = 1
C = 4
Accordingly, B2 - 4AC =
1 - 32 =
-31
Applying the quadratic formula :
-1 ± √ -31
x = ——————
4
In the set of real numbers, negative numbers do not have square roots. A new set of numbers, called complex, was invented so that negative numbers would have a square root. These numbers are written (a+b*i)
Both i and -i are the square roots of minus 1
Accordingly,√ -31 =
√ 31 • (-1) =
√ 31 • √ -1 =
± √ 31 • i
√ 31 , rounded to 4 decimal digits, is 5.5678
So now we are looking at:
x = ( -1 ± 5.568 i ) / 4
Two imaginary solutions :
x =(-1+√-31)/4=(-1+i√ 31 )/4= -0.2500+1.3919i or:
x =(-1-√-31)/4=(-1-i√ 31 )/4= -0.2500-1.3919i
Two solutions were found :
x =(-1-√-31)/4=(-1-i√ 31 )/4= -0.2500-1.3919i x =(-1+√-31)/4=(-1+i√ 31 )/4= -0.2500+1.3919i
hope it's helpful to you......
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