MHB Prove the equation has 4 distinct real solutions

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The equation \( x^3(x+1) = (x+k)(2x+k) \) is analyzed under the condition \( 0.75 < k < 1 \) to demonstrate that it has four distinct real solutions. Participants discuss methods to prove the distinctness of the solutions, emphasizing the need for a clear demonstration of how each solution differs. The conversation highlights the importance of finding explicit forms for these solutions. The mathematical exploration involves analyzing the behavior of the functions involved to confirm the existence of four distinct intersections. The discussion ultimately seeks to establish both the existence and uniqueness of these solutions within the specified range for \( k \).
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Let $0.75<k<1$.

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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anemone said:
Let $0.75<k<1$.

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

$x^3(x+1)-(x+k)(2x+k) = x^4 +x^3 -3kx -k^2 $
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} $
 
Amer said:
$x^3(x+1)-(x+k)(2x+k) = x^4 +x^3 -3kx -k^2 $
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} $

Well done, Amer! And thanks for participating...but, how are you going to show all of these four $x$ values are differ from one another?:)
 
$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
 
Hi Amer,

In order to prove that the given equation has 4 distinct real solutions, it will be enough if we prove that $\dfrac{-2+\sqrt{4-4k}}{2}<\dfrac{1-\sqrt{1+4k}}{2}$.

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

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