Yahoo Answers: Answers and Comments for How do you solve for "k" in the equation below? [Mathematics]
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From sarah M
enUS
Mon, 06 Sep 2010 00:46:25 +0000
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Yahoo Answers: Answers and Comments for How do you solve for "k" in the equation below? [Mathematics]
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From easterly: The question is looking you to construct the &...
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Sat, 10 Dec 2016 20:52:50 +0000
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

From TheSicilianSage: ... 6/(k+1)  1/(k) = 1 ← note: k ≠ 0 or 1
...
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Mon, 06 Sep 2010 00:54:18 +0000
... 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 }

From FARHAD: 5/k=0
k=infinity
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Mon, 06 Sep 2010 00:53:57 +0000
5/k=0
k=infinity