Question

The roots of $\left(x-a\right)\left(x-a-1\right)+\left(x-a-1\right)\left(x-a-2\right)+\left(x-a\right)\left(x-a-2\right)=0$, $a\in R$ are always:

• A. imaginary
• B. real and distinct
• C. equal
• D. rational and equal

Answer 1

The correct option is B

real and distinct

Explanation for the correct option:

Find the nature of roots:

To find the nature of the root we need to find Discriminant, $D=b^2-4ac$

Given that,

$\left(x-a\right)\left(x-a-1\right)+\left(x-a-1\right)\left(x-a-2\right)+\left(x-a\right)\left(x-a-2\right)=0$

Let, $\left(x-a\right)=y$, then

$$\begin{gathered} \Rightarrow y\left(y-1\right)+\left(y-1\right)\left(y-2\right)+\left(y\right)\left(y-2\right)=0 \\ \Rightarrow y^2-y+y^2-2y-y+2+y^2-2y=0 \\ \Rightarrow 3y^2-6y+2=0...\left(1\right) \end{gathered}$$

On comparing the equation $\left(1\right)$ with the standard equation $a{x}^2+bx+c=0$, we get

$$\begin{gathered} \Rightarrow a=3 \\ \Rightarrow b=-6 \\ \Rightarrow c=2 \end{gathered}$$

Substitute the value of $a,b$and $c$ in the discriminant. Then,

$\begin{array}{rcl}& \Rightarrow & D={\left(-6\right)}^2-4\left(3\right)\left(2\right)\\ & =& 36-24\\ & =& 12\end{array}$

Since,$D>0$ then roots are real and distinct.

Hence, the correct option is B.