Complex Number System!!!!!!!!!?

Everyone else is using the quadratic formula. You can also complete the square, since all the properties of algebra with real numbers also work with complex numbers. For example:

5x² + 2x + 1 = 0

x² + (2/5)x + 1/5 = 0 .... multiply both sides by 1/5, or divide by 5, your choice

[(x + 1/5)² - 1/25] + 1/5 = 0 .... because (x + 1/5)² = x² + (2/5)x + 1/25

(x + 1/5)² = -4/25 .... Simplify and then add -4/25 to each side

(x + 1/5)² = (4/25)*(-1) .... rewrite -4/25 to get the part we *can* take the square root of

x + 1/5 = ±[sqrt(4)/sqrt(25)]*sqrt(-1) ... take square roots on both sides, applying multiply/divide properties of square root on the right

x = -1/5 ± (2/5)i .... simplify square roots of 4 and 25, and substitute i for sqrt(-1)

TADAAAA!

Everything here had real coefficients, which makes things easier. If the quadratic had complex coefficients, the algebra above still works. The only minor difference would be that the square root is defined a little differently for complex numbers.

Use the quadratic formula for the equation written in the form ax^2 + bx + c = 0.

x = (-b + - sqrt (b^2 - 4ac) / 2a . . .where a = coefficient of x^2 term, b = coefficient of x term , and c = the constant. In our problem, a = 5, b = 2, and c = 1.

x = (-2 + - sqrt (2^2 - 4(5)(1)) / (2)(5)

= (-2 + - sqrt (4 - 20)) / 10

= (-2 + - sqrt (-16)) / 10

= (-2 + - sqrt ((16)(-1)) / 10

= (-2 + - 4i) / 10

= (-1 + - 2i) / 5

Use the quadratic formula where a=5, b=2 and c=1

x = -2 ±√(4) - 4(5)(1) / 2(5)

x = -1/5 ± 1/10√-16

x = -1/5 ± 2/5i

Have a good day!

a = 5

b = 2

c = 1

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

x = (- 2 +/- square_root(2*2 - 4*5*1))/(2*5)

x = (-2 +/- square_root(4 - 20))/10

x = (-2 +/- square_root(-16))/10

x = (-2 +/- 4i))/10

This last step the tricky part. square root (-16) = square_root(-1)*square_root(16)

square_root(-1) is defined is i.

square_root(16) = 4 as it always has.

x = -2/10 +/- 4i/10

x = -1/5 +/- 2*i/5

5x^2+2x+1=0

(5x^2+2x+1)/5 = 0

x^2+2/5x+1/5 = 0

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

(2/5 X 1/2)^2 ---> (2/10)^2 ---> 4/100 ---> 1/25

x^2+2/5x+1/25 = -1/5+1/25

(x+1/5)(x+1/5) = -5/25+1/25

(x+1/5)^2 = -4/25

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

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

Just use the quadratic formula for it, too

[-b±squareroot(b²-4ac)] / 2a

= [-2 ± squareroot(2² - 4*5*1)] / 2x5

= [-2 ± sq.root(-16)] / 10

= -2/10 ± (sq.root16 i) /10

= -1/5 ± (4i)/10

=- 1/5 ± 2i/5

or

= (2i-1)/5 or (-2i-1)/5

5*x^2 + 2*x + 1 = 0

sqrt(b^2 - 4*a*c) = sqrt(4 - 20) = sqrt(-16) = sqrt(16)*sqrt(-1) = ∓ 4*i

x1 =( - 2 + 4*i)/10 = ( - 1 + 2*i)/5

x1 = - 1/5 + (2*i)/5

x2 = - 1/5 - (2*i)/5

5x^2+2x+1=0

x= [-b±√ (b²-4ac)]/2a

x= [-2±√ (4-4*5*1)]/2*5

x= [-2±√ (4-4*5*1)]/10

x= [-2±√ (4-20)]/10

x= [-2±√ (-16)]/10

x= [-2±4√ (-1)]/10

x= [-2±4i]/10

x= [-1±2i]/5

use the quadratic formula:

x = [-2 +/- sqrt(4 - 4(5))] / (2*5)

x = [-2 +/- sqrt(-16)] / 10

x = [-2 +/- sqrt(16 * -1)] / 10

x = [-2 +/- sqrt(-1)sqrt(16)] / 10

x = [-2 +/- 4i] / 10 (where i = sqrt(-1))

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