- Thread starter
- #1

- Thread starter jacks
- Start date

- Thread starter
- #1

- Mar 1, 2012

- 249

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 0find value of $k$ for which the equation $x^4+(1-2k)x^2+(k^2-1) = 0$ has one real solution

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

In this instance

- $a = 1$
- $b = 1-2k$
- $c = k^2 -1$

- Jan 29, 2012

- 1,151

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.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

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

In this instance

- $a = 1$
- $b = 1-2k$
- $c = k^2 -1$

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

Last edited by a moderator:

- Moderator
- #4

- Feb 7, 2012

- 2,799

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 $k=-1.$