Question

A value of $k$ for which the quadratic equation $ {{x}^{2}}-2x(1+3k)+7(2k+3)=0 $ has equal roots, is

• A. $ 1 $
• B. $ 2 $
• C. $ 3 $
• D. $ 4 $

Answer 1

Answer: (B)

Given equation is $ {{x}^{2}}-2x(1+3k)+7(2k+3)=0. $
For equal roots, discriminant $ =0 $
$ \therefore $ $ 4{{(1+3k)}^{2}}=4\times 7(2k+3) $
$ \Rightarrow $ $ 1+9{{k}^{2}}+6k=14k+21 $
$ \Rightarrow $ $ 9{{k}^{2}}-8k-20=0 $
$ \Rightarrow $ $ (9k+10)\,(k-2)=0 $
$ \Rightarrow $ $ k=2,\,\frac{-10}{9} $