# Thread: Roots Independent of lambda.

1. ## Roots Independent of lambda.

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

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

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

(iii) $3$

(iv) $-3$

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

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

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

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

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

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

Fernando Revilla