Question

Find all values of $k$ for which there is at least one common solution of the inequality $x^2+4kx+3k^2>1+2k$ and $x^2+2kx⩽3k^2-8k+4$

Answer 1

1. $x^2+4kx+3k^2>1+2k$

$x^2+4kx+3k^2-1-2k>0$

$\Rightarrow (x-\frac{4k\pm \sqrt{16k^2+4+8k-12k^2}}{2})>0$

$\Rightarrow (x-\frac{4k\pm 2\sqrt{k^2+2k+1}}{2})>0$

$\Rightarrow x+2k\pm k+1>0$

$\Rightarrow (x+3k+1)(x+k-1)>0$ - Equation 1

2. $x^2+2kx\le 3k^2-8k+4$

$\Rightarrow x^2+2kx-3k^2+8k-4\le 0$

$\Rightarrow x-\frac{-2k\pm \sqrt{4k^2+12k^2-32k+16}}{2}$

$\Rightarrow x+k\pm 2(k-1)\le 0$

$\Rightarrow (x+3k-2)(x-k+2)\le 0$ - Equation 2

Critical points are $-3k-1,-k+1,-3k+2,k-2$

For 1, $x\in (-\infty ,-3k-1)\cup (-k+1,\infty )$

For 2, $x\in (-3k+2,k-2)$

For common solution, the above must have some interaction or they are subset of each other.

1. $\Rightarrow -3k+2<-3k-1\Rightarrow nosolution.$

2. $\Rightarrow k-2>-k+1$

$\Rightarrow 2k>3$

$\Rightarrow k>\frac{3}{2}$

$\Rightarrow k\in (\frac{3}{2},\infty )$