Find the values of 'a' for which the equation
(x2 + x + 2)2 - (a - 3)(x2 + x + 2)(x2 + x + 1) + (a - 4)(x2 + x + 1)2 = 0 has atleast one real root.
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2 Answers
1first thing you can do is expand (x2 + x + 2)2 as ((x2+x+1) + 1)2 and then treat (x2+x+1) as m or lamda whatever and then
Usual way for any quadratic equation to have real roots
Discriminant > or equal to zero!
I think You will possibly get the answer!
1708\hspace{-16}(x^2 + x + 2)^2 - (a - 3)(x^2 + x + 2)(x^2 + x + 1) + (a - 4)(x^2 + x + 1)^2 = 0\\\\ $after expanding (as myvic3 say),We Get\\\\ $(5-a).x^2+(5-a).x+(6-a)=0$\\\\ Now for al least one real Roots\;, $D\geq 0$\\\\ $(5-a)^2-4.(5-a).(6-a)\geq 0$\\\\ $(5-a).\left(5-a-24+4a\right)\geq 0$\\\\ $a\in\left[5,\frac{19}{3}\right]$\\\\ But at $a=5\;,$ We Get $1=0$(Not possible)\\\\ So $\mathbf{a\in\left(5,\frac{19}{3}\right]}$
