Question

If $(x^2+x+2{)}^2-(a-3\left)\right(x^2+x+1\left)\right(x^2+x+2)+(a-4\left)\right(x^2+x+1{)}^2=0$ has at least one real root, then the complete set of values of $a$ is

• A. $[\frac{3}{4},\infty )$
• B. $(\frac{3}{4},\infty )$
• C. $[5,\frac{19}{3})$
• D. $(5,\frac{19}{3}]$

Answer 1

The correct option is D $(5,\frac{19}{3}]$

$(x^2+x+2{)}^2-(a-3\left)\right(x^2+x+1\left)\right(x^2+x+2)+(a-4\left)\right(x^2+x+1{)}^2=0$
Dividing the equation by $x^2+x+1$ and assuming
$t=\frac{x^2+x+2}{x^2+x+1}\Rightarrow t=1+\frac{1}{x^2+x+1}\Rightarrow t=1+\frac{1}{{(x+\frac{1}{2})}^2+\frac{3}{4}}\Rightarrow t\in (1,\frac{7}{3}]$
Now, the given equation would become,
$t^2-(a-3)t+(a-4)=0\cdots \left(2\right)$
So, the given equation will have a real root when equation $\left(2\right)$ have a root in $(1,\frac{7}{3}]$
Checking
$f\left(1\right)=1-a+3+a-4=0$
Therefore, the required conditions are,
$\left(i\right)1<-\frac{b}{2a}<\frac{7}{3}\Rightarrow 1<\frac{(a-3)}{2}<\frac{7}{3}\Rightarrow 2<a-3<\frac{14}{3}\Rightarrow 5<a<\frac{23}{3}\left(ii\right)f\left(\frac{7}{3}\right)\ge 0\Rightarrow \frac{49-21(a-3)+9(a-4)}{9}\ge 0\Rightarrow 49+63-36\ge 12a\Rightarrow a\le \frac{19}{3}$
Hence, $a\in (5,\frac{19}{3}]$
(OR)
From equation $\left(2\right)$, $t^2+3t-4=a(t-1)$
$\Rightarrow (t+4)(t-1)=a(t-1)$$\Rightarrow a=t+4\because t\ne 1$
$a\in (5,\frac{19}{3}]\because t\in (1,\frac{7}{3}]$

Alternate solution:
Let $t=x^2+x+1$
We know that,
$t={(x+\frac{1}{2})}^2+\frac{3}{4}$
$\Rightarrow t\in [\frac{3}{4},\infty )$

Now,
$(t+1{)}^2-(a-3\left)t\right(t+1)+(a-4)t^2=0\Rightarrow t^2+2t+1-(a-3\left)\right(t^2+t)+a{t}^2-4t^2=0$
$\Rightarrow t(5-a)+1=0$
$\Rightarrow t=\frac{1}{a-5}$
$\Rightarrow \frac{1}{a-5}\ge \frac{3}{4}\Rightarrow \frac{1}{a-5}-\frac{3}{4}\ge 0\Rightarrow \frac{4-3(a-5)}{4(a-5)}\ge 0$
$\Rightarrow \frac{19-3a}{a-5}\ge 0$
$\Rightarrow a\in (5,\frac{19}{3}]$