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- Feb 14, 2012

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Prove the equation $x^3(x+1)=(x+k)(2x+k)$ has 4 distinct real solutions and find these solutions in explicit form.

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- #1

- Feb 14, 2012

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Prove the equation $x^3(x+1)=(x+k)(2x+k)$ has 4 distinct real solutions and find these solutions in explicit form.

Prove the equation $x^3(x+1)=(x+k)(2x+k)$ has 4 distinct real solutions and find these solutions in explicit form.

I was thinking about something like this

$(x-a)(x-b)(x-c)(x-d) = x^4 + x^3 - 2x^2 -3kx -k^2 $

but this

$(x^2 +ax + b)(x^2 + cx+ d) = x^4 + x^3 - 2x^2 -3kx -k^2 $ easier

Expanding

$x^4 + x^3(a+c) + x^2 ( b+ ac + d) + x(ad+bc) + bd = x^4 + x^3 - 2x^2 -3kx -k^2 $

we have

$ bd = -k^2 $

$ad +bc = -3k $

$b+d+ac = -2 $

$a+c = 1 $

In the third the result does not have k, and in the first b,d depend on k so there is a big possibility that $b=-d = -k $

So

Second equation

$ak - kc = -3 \Rightarrow a-c = -3 $

Last one $a+c = 1 \Rightarrow 2a = -2 , a = -1 , c =2 $

So

$(x^2 -x - k)(x^2 + 2x+ +k) = x^4 + x^3 - 2x^2 -3kx -k^2 $

$ x = \dfrac{ 1 \mp \sqrt{1 +4k}}{2} $

$ x = \dfrac{-2 \mp \sqrt{4 -4k}}{2} $

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- Feb 14, 2012

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Well done,

I was thinking about something like this

$(x-a)(x-b)(x-c)(x-d) = x^4 + x^3 - 2x^2 -3kx -k^2 $

but this

$(x^2 +ax + b)(x^2 + cx+ d) = x^4 + x^3 - 2x^2 -3kx -k^2 $ easier

Expanding

$x^4 + x^3(a+c) + x^2 ( b+ ac + d) + x(ad+bc) + bd = x^4 + x^3 - 2x^2 -3kx -k^2 $

we have

$ bd = -k^2 $

$ad +bc = -3k $

$b+d+ac = -2 $

$a+c = 1 $

In the third the result does not have k, and in the first b,d depend on k so there is a big possibility that $b=-d = -k $

So

Second equation

$ak - kc = -3 \Rightarrow a-c = -3 $

Last one $a+c = 1 \Rightarrow 2a = -2 , a = -1 , c =2 $

So

$(x^2 -x - k)(x^2 + 2x+ +k) = x^4 + x^3 - 2x^2 -3kx -k^2 $

$ x = \dfrac{ 1 \mp \sqrt{1 +4k}}{2} $

$ x = \dfrac{-2 \mp \sqrt{4 -4k}}{2} $

$f(x) = x^4 +x^3 -3kx -k^2 $

$ f(x) = x^3(x+1) - (x+k)(2x+k) $

For

$ 0.75 < k < 1 $

$ f(1) = 2 - 3k-k^2 < 0 $ for the any k in the range

$f(2) = 8(3) - (2+k)(4+k) > 0 $

we have a root in (1,2)

$f(-1) > 0 , f(-2) < 0 $

we have a root in (-2,-1)

Not sure about this one

$ f(0) < 0 , f(-0.5) >0 $

root at (-0.5 , 0)

still one

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- #5

- Feb 14, 2012

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From $0.75<k<1$, we get that $-2<\dfrac{-2+\sqrt{4-4k}}{2}<-1.5$ and $-0.618<\dfrac{1-\sqrt{1+4k}}{2}<-0.5$.

Hence we can conclude that $\dfrac{-2+\sqrt{4-4k}}{2}<\dfrac{1-\sqrt{1+4k}}{2}$ which in turn gives us

$\dfrac{-2-\sqrt{4-4k}}{2}<\dfrac{-2+\sqrt{4-4k}}{2}<\dfrac{1-\sqrt{1+4k}}{2}<\dfrac{1+\sqrt{1+4k}}{2}$