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)

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).