Matematické Fórum

stuart clark · 26. 10. 2011 04:24

Find the values of $a$ for which the equation

$(x^2 + x + 2)^2 - (a - 3)(x^2 + x + 2)(x^2 + x + 1) + (a - 4)(x^2 + x + 1)^2 = 0$ has atleast one real root.

stuart clark · 27. 10. 2011 15:57

I have solved in this way

Now here $x^2+x+1> 0\forall x\in \mathbb{R}\;,$ so We can divide by $x^2+x+1$

$\left(\frac{x^2+x+2}{x^2+x+1}\right)^2-(a-3).\left(\frac{x^2+x+2}{x^2+x+1}\right)+(a-4)=0$

now Let $\left(\frac{x^2+x+2}{x^2+x+1}\right)=t$

where $1<t \leq \frac{7}{3}$

So $t^2-(a-3).t+(a-4)=0$

for at least one real roots. $D\geq 0$

so $(a-3)^2-4.(a-4)\geq 0$

$(a-5)^2\geq 0\Leftrightarrow a\in\mathbb{R}$

Now from here what can i do..

help required.