Question

If one root of the equation $ {{x}^{2}}+px+12=0 $ is 4 when equation $ {{x}^{2}}+px+q=0 $ have equal roots, then the value of $ q $ is

• A. $ \frac{4}{49} $
• B. $ \frac{49}{4} $
• C. $ 4 $
• D. $ \frac{1}{4} $

Answer 1

Answer: (B)

Given, 4 is a root of the equation
$ {{x}^{2}}+px+12=0 $
$ \therefore $ $ 16+4p+12=0 $
$ \Rightarrow $ $ 4p=-28 $
$ \Rightarrow $ $ p=-7 $
And equation $ {{x}^{2}}+px+q=0 $
have equal roots, let roots are
$ \alpha ,\alpha $ .
$ \therefore $ $ \alpha +\alpha =-\frac{p}{1}=-p $
$ \Rightarrow $ $ 2\alpha =-p $
$ \Rightarrow $ $ \alpha =-\frac{p}{2}=\frac{7}{2} $
$ [\because p=-7] $ and $ \alpha .\alpha =q $
$ \Rightarrow $ $ {{\left( \frac{7}{2} \right)}^{2}}=q $
$ \Rightarrow $ $ q=\frac{49}{4} $