Welcome to our community

Be a part of something great, join today!

Prove the equation has 4 distinct real solutions

  • Thread starter
  • Admin
  • #1

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,678
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.
 

Amer

Active member
Mar 1, 2012
275
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} $
 
  • Thread starter
  • Admin
  • #3

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,678
$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?:)
 

Amer

Active member
Mar 1, 2012
275

$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


 
  • Thread starter
  • Admin
  • #5

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,678
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

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