Question

Consider the equation $x^2+2(a-1)x+a+5=0$, where $a$ is a parameter. Match the real values of $a$ so that the given equation has

Answer 1

$x^2+2(a-1)x+a+5=0$
Here, $D=4(a-1{)}^2-4(a+5)=4(a+1\left)\right(a-4)$
(i) For imaginary roots
$D<0$
$\Rightarrow (a+1)(a-4)<0$
$\Rightarrow a\in (-1,4)$
(ii) One root smaller than $3$ and other root greater than $3$ means $3$ lies in between the roots .
$D>0$ and $af\left(3\right)<0$
$\Rightarrow (a+1)(a-4)>0$ and $9+6(a-1)+a+5<0$
$\Rightarrow a\in (-\infty ,-1)\cup (4,\infty )$ and $8+7a<0$
$\Rightarrow a\in (-\infty ,-1)\cup (4,\infty )$ and $a\in (-\infty ,-\frac{8}{7})$
$\Rightarrow a\in (-\infty ,-\frac{8}{7})$
(iii) For this case,
$D\ge 0$ and $f\left(1\right)f\left(3\right)<0$
$\Rightarrow (a+1)(a-4)\ge 0$ and $(3a+4)(7a+8)<0$
$\Rightarrow a\in (-\infty ,-1]\cup [4,\infty )$ and $a\in (-\frac{4}{3},-\frac{8}{7})$
$\Rightarrow a\in (-\frac{4}{3},-\frac{8}{7})$
(iv) For this case
$D>0$ and $f\left(1\right)<0$
$\Rightarrow (a+1)(a-4)>0$ and $3a+4<0$
$\Rightarrow a\in (-\infty ,-1)\cup (4,\infty )$ and $a<-\frac{4}{3}$
$\Rightarrow a\in (-\infty ,-\frac{4}{3})$