Question

If$\alpha$ and$\beta$ are the roots of the equation$a{x}^2+bx+c=0$, $\alpha \beta =3$ and$a,bandc$ are in AP, then$\alpha +\beta$ is equal to

• A. $-4$
• B. $1$
• C. $4$
• D. $-2$

Answer 1

The correct option is D

$-2$

Explanation for correct option:

Step 1: Apply the A.P relation between three given numbers$\mathit{a}\mathbf{,}\mathbf{}\mathit{b}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathit{c}$.

We have given a quadratic equation$a{x}^2+bx+c=0$, whose roots are$\alpha$ and$\beta$, $\alpha \beta =3$and $a,bandc$ are in A.P.

We have to find the value of$\alpha +\beta$.

As$a,bandc$ are in A.P.

$\therefore 2b=a+c$

Step 2: Find the relationship$\mathit{a}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathit{c}$.

and we have$\alpha \beta =3$

$$\begin{gathered} \alpha \beta =\frac{c}{a} \\ \Rightarrow 3a=c \end{gathered}$$

Step 3: Find the relationship$\mathit{a}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathit{b}$.

so, above first equation.

$$\begin{gathered} 2b=a+3a \\ \Rightarrow 2b=4a \\ \Rightarrow b=2a \end{gathered}$$

Step 4: Finding the value of the sum of roots.

$$\begin{gathered} \alpha +\beta =\frac{-b}{a} \\ \Rightarrow \alpha +\beta =\frac{-2a}{a}=-2 \\ \Rightarrow \alpha +\beta =-2 \end{gathered}$$

Hence, the correct option is$\left(D\right)$.