Question

The equation $(a+2)x^2+(a-3)x-(2a-1)=0$ has rational roots for

• A. $a\in R-\{-2\}$
• B. $a\in Q-\{-2\}$
• C. $a\in (-2,\infty )$
• D. $a\in Q$

Answer 1

The correct option is B $a\in Q-\{-2\}$

For the equation to remain quadratic, the first term should not be $0$.
Hence, $a$ cannot take the value $-2$ .
According to the question, $\Delta$ has to be rational since the roots are rational numbers and $a,b,c$ are also rational .
Here, $\Delta =(a-3{)}^2+4(a+2\left)\right(2a-1)=9a^2+6a+1=(3a+1{)}^2$
If $(3a+1)$ is rational, then the roots would also be rational .
Hence, Option B is correct .