Question

.find the value of p such that the quadratic equation 4x² - 2(p+1)x + (p+4) = 0 has equal roots

Answer 1

$$\begin{gathered} 4{\mathrm{x}}^2-2\left(\mathrm{p}+1\right)\mathrm{x}+\left(\mathrm{p}+4\right)=0 \\ \mathrm{Now},\mathrm{a}=4;\mathrm{b}=-2\left(\mathrm{p}+1\right);\mathrm{c}=\left(\mathrm{p}+4\right) \\ \mathrm{Now},\mathrm{D}={\mathrm{b}}^2-4\mathrm{ac} \\ ={\left[-2\left(\mathrm{p}+1\right)\right]}^2-4\left(4\right)\left(\mathrm{p}+4\right) \\ =4\left({\mathrm{p}}^2+1+2\mathrm{p}\right)-16\mathrm{p}-64 \\ =4\left[{\mathrm{p}}^2+1+2\mathrm{p}-4\mathrm{p}-16\right] \\ =4\left({\mathrm{p}}^2-2\mathrm{p}-15\right) \\ \mathrm{For}\mathrm{equal}\mathrm{roots}, \\ \mathrm{D}=0 \\ \Rightarrow 4\left({\mathrm{p}}^2-2\mathrm{p}-15\right)=0 \\ \Rightarrow {\mathrm{p}}^2-2\mathrm{p}-15=0 \\ \Rightarrow {\mathrm{p}}^2-5\mathrm{p}+3\mathrm{p}-15=0 \\ \Rightarrow \mathrm{p}\left(\mathrm{p}-5\right)+3\left(\mathrm{p}-5\right)=0 \\ \Rightarrow \left(\mathrm{p}+3\right)\left(\mathrm{p}-5\right)=0 \\ \Rightarrow \mathrm{p}=-3;\mathrm{p}=5 \end{gathered}$$