# Thread: Roots Independent of lambda.

1. ## Roots Independent of lambda.

If $\displaystyle x^5+5\lambda x^4+(\lambda a-4)x^2-(8\lambda+3)x+\lambda a-2=0$. Then the value of $\displaystyle a$ for which the Given equation has one root Independent of $\displaystyle \lambda$

(i) $\displaystyle \frac{64}{5}$

(ii)$\displaystyle -\frac{64}{5}$

(iii) $\displaystyle 3$

(iv)$\displaystyle -3$

2. Pick a value for lambda (1,2,3,...). Since this is multiple choice, you should be able to eliminate the wrong choices (plot the graphs).

3. Originally Posted by jacks
$\displaystyle x^5+5\lambda x^4+(\lambda a-4)x^2-(8\lambda+3)x+\lambda a-2=0$. Then the value of $\displaystyle a$ for which the Given equation has one root Independent of $\displaystyle \lambda$
In general, for a family of polynomials:

$\displaystyle \{f_{\lambda,\;a}(x)=p(x)+\lambda q_a(x)\}$

if $\displaystyle r$ is a root of $\displaystyle p(x)$ and $\displaystyle q_a(x)$ then, $\displaystyle r$ is a root of $\displaystyle f_{\lambda,\;a}(x)$. So, finding a root $\displaystyle r$ of $\displaystyle p(x)$ and substituting: $\displaystyle q_a(r)=0$ , we obtain a sufficient condition for $\displaystyle a$.

That is a pity, if $\displaystyle r=2$ were a root of $\displaystyle p(x)$ (it isn't) then,

$\displaystyle q_a(2)=0 \Leftrightarrow a=-64/5$

Possiibly (of course I'm not sure) there is typo.

Fernando Revilla