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

5x^2+2x+1=0

how do i solve this using the complex number system.

### 9 Answers

- husoskiLv 71 decade agoBest Answer
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.

- HPVLv 71 decade ago
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

- DonLv 41 decade ago
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!

- jcherry_99Lv 71 decade ago
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

- How do you think about the answers? You can sign in to vote the answer.
- 1 decade ago
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

- 1 decade ago
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

- GuillermoLv 71 decade ago
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

- momLv 71 decade ago
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

- notthejakeLv 71 decade ago
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