Question

If the equation $(a+1)x^2-(a+2)x+(a+3)=0$ has roots equal in magnitude but opposite in sign, then the roots of the equation are

• A. $\pm 1$
• B. $\pm \frac{a}{2}$
• C. $\pm \frac{3a}{2}$
• D. $\pm 2a$

Answer 1

The correct option is B

$\pm \frac{a}{2}$

Explanation for correct option

Step 1: Solve for the value of $a$

We know that if $\alpha ,\beta$ are the roots of a quadratic equation $p{x}^2+qx+r=0$ then

$\alpha +\beta =-\frac{q}{p}$ …..$\left(i\right)$

$\alpha \beta =\frac{r}{p}$ …..$\left(ii\right)$

Given quadratic equation is $(a+1)x^2-(a+2)x+(a+3)=0$

Now comparing with the general form of quadratic equation we have $p=a+1$, $q=-\left(a+2\right)$ and $r=a+3$

Given that the roots are of same magnitude and opposite sign

Let the root are $\alpha ,-\alpha$

From $\left(i\right)$

$$\begin{gathered} \therefore \alpha -\alpha =\frac{a+2}{a+1} \\ \Rightarrow a+2=0 \\ \Rightarrow a=-2 \end{gathered}$$

Step 2: Solve for the roots of the quadratic equation

From$\left(ii\right)$

$$\begin{gathered} \therefore -{\alpha }^2=\frac{a+3}{a+1} \\ \Rightarrow -{\alpha }^2=\frac{-2+3}{-2+1} \\ \Rightarrow {\alpha }^2=1 \\ \Rightarrow \alpha =\pm 1 \end{gathered}$$

Clearly the roots are $\pm \frac{a}{2}\left[\because a=-2\right]$

Hence, option (B) is correct i.e. $\pm \frac{a}{2}$