x^2+6x-4=0. Solve with exact answers.?

Type exact answer, if you need to use radicals you can. Separate answers with commas.

6 Answers

Relevance
  • Dave
    Lv 5
    9 years ago
    Favorite Answer

    x = (-b + sqrt(b^2 - 4ac)) / (2a)

    x = (-6 + sqrt(6^2 - 4(1)(-4)) / (2(1))

    x = (-6 + sqrt(36 - (-16))) / 2

    x = (-6 + sqrt(36 + 16)) / 2

    x = (-6 + sqrt(52)) / 2

    x = (-6 + 2sqrt(13)) / 2

    x = -3 + sqrt(13)

    x = (-b - sqrt(b^2 - 4ac)) / (2a)

    x = (-6 - sqrt(6^2 - 4(1)(-4)) / (2(1))

    x = (-6 - sqrt(36 - (-16))) / 2

    x = (-6 - sqrt(36 + 16)) / 2

    x = (-6 - sqrt(52)) / 2

    x = (-6 - 2sqrt(13)) / 2

    x = -3 - sqrt(13)

    So x equals:

    -3 + sqrt(13)

    and

    -3 - sqrt(13)

  • Anonymous
    9 years ago

    easy peasy

  • 9 years ago

    x^2 + 6x - 4 = 0

    Use quadratic formula

    x= -b +_ square root of (b^2 - 4 ac)

    -----------------------------------------------

    2a

    = - 6 +_ sqrt( (6)^2 - 4(1) (-4) )

    ------------------------------------------

    2(1)

    = -6 +_ sqrt(36+16)

    --------------------------

    2

    = - 6 +_ sqrt(52)

    --------------------

    2

    = -6 + 2 +_ (sqrt) 13 (The 52 is written as simpler. (2 root of 13)

    ------------------------------

    2

    = -3 +_ sqrt(13) -----> answer

    Hope it helps!

    Source(s): C$K
  • 9 years ago

    Quadratic formula: (-b±sqrt(b^2-4ac))/2a

    a=1, b=6, c=-4

    Therefore...

    (-6±sqrt(6^2-4*1*(-4)))/(2*1) = (-6±sqrt(36+16))/2 = (-6±sqrt(52))/2

    sqrt(52) can be written as 2*sqrt(13).

    (-6±2*sqrt(13))/2=-3±sqrt(13)

    And that's your answer. -3±sqrt(13)

  • How do you think about the answers? You can sign in to vote the answer.
  • Anonymous
    9 years ago

    x²+6x-4 = 0

    = (x²+6x+9)-13 = 0

    = (x+3)²-(√13)² = 0

    = (x+3+√13)(x+3-√13) = 0

    As either of the multiplier must be 0, then x = -3+√13, -3-√13

  • 9 years ago

    this is why we have a quadratic formula

Still have questions? Get your answers by asking now.