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

5x^2+2x+1=0

how do i solve this using the complex number system.

9 Answers

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  • 1 decade ago
    Best 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.

  • HPV
    Lv 7
    1 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

  • Don
    Lv 4
    1 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!

  • 1 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

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  • 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

  • 1 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

  • mom
    Lv 7
    1 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

  • 1 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

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