Question

​​given that 2 is a root of the equation 3(x)(sq.)-p(x+1)=0 & that the equation p(x)(sq.)-qx+9=0 has equal roots,find the value of p & q.
IT IS 31st QUES. OF EX.5B QUADRATIC EQUATIONS -CONCISE MATHS SELINA

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{given}\mathrm{quadratic}\mathrm{equation}\mathrm{is}, \\ 3{\mathrm{x}}^2-\mathrm{p}\left(\mathrm{x}+1\right)=0 \\ \mathrm{Since},2\mathrm{is}\mathrm{the}\mathrm{root}\mathrm{of}\mathrm{the}\mathrm{above}\mathrm{equation},\mathrm{then}\mathrm{it}\mathrm{must}\mathrm{satisfy}\mathrm{it}. \\ \mathrm{Now},3{\left(2\right)}^2-\mathrm{p}\left(2+1\right)=0 \\ \Rightarrow 12-3\mathrm{p}=0 \\ \Rightarrow 3\mathrm{p}=12 \\ \Rightarrow \mathrm{p}=4 \\ \mathrm{Now},\mathrm{the}\mathrm{other}\mathrm{quadratic}\mathrm{equation}\mathrm{is}, \\ {\mathrm{px}}^2-\mathrm{qx}+9=0 \\ \Rightarrow 4{\mathrm{x}}^2-\mathrm{qx}+9=0\left[\mathrm{On}\mathrm{putting}\mathrm{p}=4\right] \\ \mathrm{Now},\mathrm{a}=4;\mathrm{b}=-\mathrm{q};\mathrm{c}=9 \\ \mathrm{Now},\mathrm{D}={\mathrm{b}}^2-4\mathrm{ac}={\left(-\mathrm{q}\right)}^2-4\left(4\right)\left(9\right)={\mathrm{q}}^2-144 \\ \mathrm{Now},\mathrm{for}\mathrm{equal}\mathrm{roots}, \\ \mathrm{D}=0 \\ \Rightarrow {\mathrm{q}}^2-144=0 \\ \Rightarrow {\mathrm{q}}^2=144 \\ \Rightarrow \mathrm{q}=\pm 12 \end{gathered}$$