Question

If x = -2 is a root of the equation 3x²+ 7x + p= 0, find the value of 'k' so that the roots of the equation x² + k( 4x + k- 1)+p = 0 are equal.

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{given}\mathrm{quadratic}\mathrm{equation}\mathrm{is}, \\ 3x^2+7x+p=0 \\ \mathrm{Since},x=-2\mathrm{is}\mathrm{a}\mathrm{root}\mathrm{of}\mathrm{the}\mathrm{above}\mathrm{equation},\mathrm{so}\mathrm{it}\mathrm{must}\mathrm{satisfy}\mathrm{it}. \\ \mathrm{Put}x=-2\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{equation},\mathrm{we}\mathrm{get} \\ 3{\left(-2\right)}^2+7\left(-2\right)+p=0 \\ \Rightarrow 12-14+p=0 \\ \Rightarrow p=2 \\ \mathrm{Now},\mathrm{the}\mathrm{given}\mathrm{quadratic}\mathrm{equation}\mathrm{is}, \\ {x}^2+k\left[4x+k-1\right]+p=0 \\ \Rightarrow x^2+4kx+\left(k^2-k+2\right)=0 \\ \mathrm{Now},a=1;b=4k;c=\left(k^2-k+2\right) \\ \mathrm{Now},\mathrm{D}=b^2-4ac={\left(4k\right)}^2-4\left(1\right)\left(k^2-k+2\right)=16k^2-4k^2+4k-8 \\ \mathrm{So},\mathrm{D}=12k^2+4k-8 \\ \mathrm{Now},\mathrm{the}\mathrm{above}\mathrm{equation}\mathrm{has}\mathrm{equal}\mathrm{roots},\mathrm{then} \\ \mathrm{D}=0 \\ \Rightarrow 12k^2+4k-8=0 \\ \Rightarrow 3k^2+k-2=0 \\ \Rightarrow 3k^2+3k-2k-2=0 \\ \Rightarrow 3k\left(k+1\right)-2\left(k+1\right)=0 \\ \Rightarrow \left(3k-2\right)\left(k+1\right)=0 \\ \Rightarrow 3k-2=0\mathrm{or}k+1=0 \\ \Rightarrow k=\frac{2}{3}\mathrm{or}k=-1 \end{gathered}$$