Question

If the equation $x^2-2x(1+3k)+7(3+2k)=0$ has equal roots, then k = ?
(a) $2\mathrm{or}\frac{10}{9}$
(b) $-2\mathrm{or}\frac{10}{9}$
(c) $2\mathrm{or}\frac{-10}{9}$
(d) $-2\mathrm{or}\frac{-10}{9}$

Answer 1

(c) $2\mathrm{or}\frac{-10}{9}$

$$\begin{gathered} \text{It is given that the roots of the equation (}{x}^2-2x(1+3k)+7(3+2k)=0\text{are equal}\text{.} \\ \therefore (b^2-4ac)=0 \\ \Rightarrow {\left\{2(1+3k)\right\}}^2-4\times 1\times 7(3+2k)=0 \\ \Rightarrow 4(9k^2+6k+1)-28(3+2k)=0 \\ \Rightarrow 36k^2+24k+4-84-56k=0 \\ \Rightarrow 36k^2-32k-80=0 \\ \Rightarrow 9k^2-8k-20=0 \\ \Rightarrow 9k^2-18k+10k-20=0 \\ \Rightarrow 9k(k-2)+10(k-2)=0 \\ \Rightarrow (k-2)(9k+10)=0 \\ \Rightarrow k=2\text{or}k=\frac{-10}{9} \end{gathered}$$