how do you solve for "k" in the equation below?

(6/k+1)-(1/k)=1

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  • 9 years ago
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    ... 6/(k+1) - 1/(k) = 1 ← note: k ≠ 0 or -1

    or 6(k) - (k+1) = (k+1)(k)

    or 6k - k - 1 = k² + k

    or 0 = k² + k- 6k + k + 1

    or 0 = k² - 4k + 1

    use the quadratic formula:

    k = [ -b ± √ ( (b)² - 4(a)(c) ) ] / 2(a)

    k = [ -(-4) ± √ ( (-4)² - 4(1)(1) ) ] / 2(1)

    k = [ 4 ± √ ( 16 - 4 ) ] / 2

    k = [ 4 ± √ ( 12 ) ] / 2

    k = [ 4 ± √ ( 4*3 ) ] / 2

    k = [ 4 ± 2√3 ) ] / 2

    k = 2 ± √3

    k = { 2 - √3, 2 + √3 }

    k ~ { 0.2679, 3.7321 }

  • 3 years ago

    The question is looking you to construct the "quadratic formula" from scratch. The quadratic formula is one recipe to locate values of x in a quadratic equation. polishing off the sq. is a distinctive recipe. considering the fact that they the two supply the comparable solutions, it will be accessible to bypass from one to the different. A sq. has the style: (x+ok)^2 = x^2 + 2kx + ok^2 As you will locate, that's ultimate proper for "monic" quadratics (fancy be conscious to advise that the 1st coefficient is a a million) So, first step, divide your comprehensive equation by making use of "a" (a/a)x^2 + (b/a)x + (c/a) = 0/a a/a is the comparable as a million and 0/a is the comparable as 0 x^2 + (b/a)x + (c/a) = 0 that's now a "monic" quadratic and it ought to have the comparable x values with the aid of fact the unique one (with the aid of undeniable fact that's equivalent) next, we ought to stress it to look like the sq.: x^2 + 2kx + ok^2 so we ought to make 2k = b/a this forces us to apply ok = b/2a which then provides us ok^2 = (b/2a)^2 = (b^2) / 4a^2 however the equation has a "c/a" instead of the value of ok^2 so we pass the c/a to the different area x^2 + (b/a)x + 0 = -(c/a) and upload [b^2 / 4a^2] to the two factors x^2 + (b/a)x + b^2/4a^2 = b^2/4a^2 - c/a The left area first: all of us comprehend the left area is a sq. with the aid of fact we in simple terms busted our derrieres to make it so. (x + (b/2a))^2 = b^2/4a^2 - c/a next, the main marvelous area. First, positioned each and every thing over the comparable denominator (4a^2) (x + (b/2a))^2 = (b^2 - 4ac)/4a^2 Now, sq. root the two factors, remembering tha the effects of a sq. root could be advantageous or adverse; to illustrate, the sq. root of +4 must be -2, with the aid of fact (-2)^2 = +4. x + b/2a = +/- sqrt(b^2 - 4ac) / 2a +/- potential "plus or minus" pass the b/2a to the main marvelous: x = -b/2a +/- sqrt(b^2 - 4ac)/2a ingredient out the denominator (that's the comparable 2a for the two words) x = [ -b +/- sqrt( b^2 - 4ac ) ] / 2a

  • 9 years ago

    5/k=0

    k=infinity

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