Question

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

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

Answer 1

Answer: (B)

$\pm \,\frac{1}{2}\,a$

Answer 2

Given equation is $(a+1){{x}^{2}}-(a+2)x+(a+3)=0$
Since, roots are equal in magnitude and opposite in sign.
$\therefore$ coefficient of x is zero ie, $a+2=0$
$\Rightarrow$ $a=-2$ ..(i)
$\therefore$ Equation is $(-2+1){{x}^{2}}-(-2+2)x+(-2+3)=0$
$\Rightarrow$ $-{{x}^{2}}+1=0$
$\Rightarrow$ $x=\pm 1$ ..(ii)
Only option [b] ie, $\pm \frac{1}{2}$
a satisfies Eqs. (i) and (ii).