Question

If equations $2x^2+3x+4=0$ and $4a{x}^2+5bx+6c=0$ (where $a,b,c\in N$ ) have a common root then minimum value of $(a+b+c)$ is equal to

• A. 9
• B. 15
• C. 53
• D. 47

Answer 1

Answer: (C)

$\Theta$ Roots of $1^{\text{st }}$ quadratic are imaginary
$\therefore$ Both roots will be common.
$\therefore \frac{4a}{2}=\frac{5b}{3}=\frac{6c}{4}=k$
$a=\frac{k}{2},b=\frac{3k}{5},c=\frac{2k}{3}$
For $a,b,c$ to be natural number minimum value of $k$ is LCM of 2,3,5 i.e. 30
$\therefore a=15,b=18,c=20$
$\therefore a+b+c=53$