Question

If abx² = (a-b)²(x+1)

Then the possible value of 1+ 4/x +1/4x² is ?

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{given}\mathrm{quadratic}\mathrm{equation}\mathrm{is} \\ ab{x}^2={\left(a-b\right)}^2\left(x+1\right) \\ \Rightarrow ab{x}^2-{\left(a-b\right)}^{2}x-{\left(a-b\right)}^2=0 \\ \mathrm{We}\mathrm{will}\mathrm{solve}\mathrm{the}\mathrm{above}\mathrm{equation}\mathrm{using}\mathrm{quadratic}\mathrm{formula}. \\ \mathrm{Comparing}\mathrm{it}\mathrm{with}\mathrm{standard}\mathrm{quadratic}\mathrm{equation}\mathrm{A}{x}^2+\mathrm{B}x+\mathrm{C}=0,\mathrm{we}\mathrm{get} \\ \mathrm{A}=ab;\mathrm{B}=-{\left(a-b\right)}^2;\mathrm{C}=-{\left(a-b\right)}^2 \\ \mathrm{Discriminant},\mathrm{D}={\mathrm{B}}^2-4\mathrm{AC} \\ \mathrm{D}={\left(a-b\right)}^4+4ab{\left(a-b\right)}^2 \\ \mathrm{D}={\left(a-b\right)}^2\left[{\left(a-b\right)}^2+4ab\right] \\ \mathrm{D}={\left(a-b\right)}^2\left[a^2+b^2-2ab+4ab\right] \\ \mathrm{D}={\left(a-b\right)}^2\left(a^2+b^2+2ab\right) \\ \mathrm{D}={\left(a-b\right)}^2{\left(a+b\right)}^2 \end{gathered}$$

$$\begin{gathered} \mathrm{Now},\mathrm{values}\mathrm{of}x\mathrm{are}\mathrm{given}\mathrm{by}, \\ x=\frac{-\mathrm{B}+\sqrt{\mathrm{D}}}{2\mathrm{A}}\mathrm{and}x=\frac{-\mathrm{B}-\sqrt{\mathrm{D}}}{2\mathrm{A}} \\ x=\frac{{\left(a-b\right)}^2+\left(a-b\right)\left(a+b\right)}{2ab}\mathrm{and}x=\frac{{\left(a-b\right)}^2-\left(a-b\right)\left(a+b\right)}{2ab} \\ x=\frac{\left(a-b\right)\left[a-b+a+b\right]}{2ab}\mathrm{and}x=\frac{\left(a-b\right)\left[a-b-a-b\right]}{2ab} \\ x=\frac{\left(a-b\right)\left(2a\right)}{2ab}=\frac{a-b}{b}\mathrm{and}x=\frac{\left(a-b\right)\left(-2b\right)}{2ab}=\frac{b-a}{a} \\ \mathrm{Now}\mathrm{you}\mathrm{can}\mathrm{easily}\mathrm{get}\mathrm{the}\mathrm{values}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{expression}\mathrm{by}\mathrm{substituting} \\ \mathrm{both}\mathrm{the}\mathrm{values}\mathrm{of}x\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{expression}. \end{gathered}$$