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one real solution

jacks

Well-known member
Apr 5, 2012
226
find value of $k$ for which the equation $x^4+(1-2k)x^2+(k^2-1) = 0$ has one real solution
 
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SuperSonic4

Well-known member
MHB Math Helper
Mar 1, 2012
249
find value of $k$ for which the equation $x^4+(1-2k)x^2+(k^2-1) = 0$ has one real solution
This is a quadratic equation (let $u=x^2$ if you need to make it clearer) so you can use the discriminant to find the values of k - we know that there is one real solution if the discriminant is equal to 0

Hence we're looking for a value of k such that: $b^2 - 4ac = 0$

In this instance
  • $a = 1$
  • $b = 1-2k$
  • $c = k^2 -1$
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
This is a quadratic equation (let $u=x^2$ if you need to make it clearer) so you can use the discriminant to find the values of k - we know that there is one real solution if the discriminant is equal to 0
I don't believe that is correct. If the discriminant is 0, then the quadratic has one real root. But if that root is negative, there is no real solution to the quartic while if it is positive, there are two real roots. In order that there be only one real root to the original quartic is if the quadratic has, as its only positive root, u= 0 so that the only real root to the original quartic is 0. That means that the quadratic in u must be simply [tex]u^2= 0[/tex]. In other words, we must have both 1- 2k= 0 and [tex]k^2- 1= 0[/tex]. The first equation has only k= 1/2 as solution and the second has only k= 1 or k= -1 as solutions. There is no value of k that makes both 0 so there is no value of k that gives only a single real root to the original quartic equation.

Hence we're looking for a value of k such that: $b^2 - 4ac = 0$

In this instance
  • $a = 1$
  • $b = 1-2k$
  • $c = k^2 -1$
With those equations,
b^2= (1- 2k)^2= 4k^2- 4k+ 1
4ac= 4k^2- 4 so b^2- 4ac= 4k^2- 4k+ 1- 4k^32+ 4= -4k+ 5= 0 so k= 5/4.
In that case, 1- 2k= 1- 5/2= -3/2 and k^2- 1= 25/16- 1= 9/16. The quadratic equation for u is [tex]u^2- (3/2)u+ 9/16= (u- 3/4)^2= 0[/tex] so that [tex]u= 3/4[/tex]. But then [tex]x^2= 3/4[/tex] so the original equation has the two roots [tex]x= \sqrt{3}/2[/tex] and [tex]-\sqrt{3}/2[/tex]
 
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Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,708
We are slowly getting there! The equation is a quadratic in $x^2$. Therefore if $x_0$ is a solution, so is $-x_0$. The only way for there to be a unique real solution is if that solution is 0. The condition for 0 to be a solution is that the constant term vanishes. Thus $k^2-1=0,$ and so $k= 1$ or $k=-1.$ If $k=1$ then the equation becomes $x^4 -x^2=0$. But that has solutions $x=\pm1$ as well as $x=0.$ So we can rule out that case, and the only remaining possibility is $k=-1.$

If $k=-1$ then the equation becomes $x^4+3x^2=0$, and that does indeed have $x=0$ as its only real root. So the answer to the problem is that $k=-1.$