Question

If $a{x}^2+bx+c=0$ and $2x^2+3x+4=0$ have a common root, where $a,b,c\in N$(set of natural numbers), then the least value of $a+b+c$ is

• A. $13$
• B. $11$
• C. $7$
• D. $9$

Answer 1

The correct option is D

$9$

Find the least value of $\mathit{a}\mathbf{}\mathbf{+}\mathbf{}\mathit{b}\mathbf{}\mathbf{+}\mathbf{}\mathit{c}$:

$2x^2+3x+4=0$

$$\begin{gathered} x=\frac{-3\pm \sqrt{9-4\times 2\times 4}}{2\times 2} \\ x=\frac{-3\pm \sqrt{9-32}}{4} \end{gathered}$$

$\therefore$Roots of the quadratic are complex.

$\therefore$ Both roots are common in the equations $2x^2+3x+4=0$ and$a{x}^2+bx+c=0$
Because complex roots occur in pairs.

$\therefore \frac{a}{2}=\frac{b}{3}=\frac{c}{4}$

then the least value of $a+b+c$ is $2+3+4=9$

Hence, the correct option is D.