Question

If equations $x^2-3x+4=0$ and $4x^2-2[3a+b]x+b=0(a,b\in R)$ have a common root, then the complete set of values of $a$ is
$($Here, $\left[K\right]$ denotes the greatest integer less than or equal to $K.)$

• A. $[\frac{-10}{3},-3)$
• B. $(3,\frac{10}{3}]$
• C. $[\frac{-11}{3},\frac{-10}{3})$
• D. $(\frac{-10}{3},-3]$

Answer 1

The correct option is A $[\frac{-10}{3},-3)$

$x^2-3x+4=0$
$D=(-3{)}^2-4\times 1\times 4=-7<0$
$\because D<0,$ hence $x^2-3x+4=0$ have imaginary roots.
$\Rightarrow$ Both roots of the given equations are common.
So, $\frac{4}{1}=\frac{-2[3a+b]}{-3}=\frac{b}{4}$
$\text{(I)}\text{(II)}\text{(III)}$

Taking $\text{(I)}$ and $\text{(III)},$ we get
$b=16$
Taking $\text{(I)}$ and $\text{(II)},$ we get
$[3a+b]=6\cdots \left(1\right)$
Putting the value of $b$ in equation $\left(1\right),$ we get
$[3a+16]=6$
$\Rightarrow \left[3a\right]+16=6$
$\Rightarrow \left[3a\right]=-10$
$\Rightarrow -10\le 3a<-9$
$\Rightarrow \frac{-10}{3}\le a<-3$