Question

$16x^2=24x+1$

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ 16x^2=24x+1 \\ \text{⇒ 16}{x}^2-24x-1=0 \\ \mathrm{On}\mathrm{c}\text{omparing it with}a{x}^2+bx+c=0 \\ a=16,b=-24\mathrm{and}c=-1 \\ \mathrm{Discriminant}D\mathrm{is}\mathrm{given}\mathrm{by}: \\ D=(b^2-4ac) \\ \text{=}{(-24)}^2-4\times 16\times (-1) \\ \text{= 576}+\left(64\right) \\ \text{= 640}>0 \\ \text{Hence, the roots of the equation are real,} \\ \mathrm{Roots}\alpha \mathrm{and}\beta \mathrm{are}\mathrm{given}\mathrm{by}: \\ \alpha =\frac{-b+\sqrt{D}}{2a}\text{=}\frac{-(-24)+\sqrt{640}}{2\times 16}\text{=}\frac{24+8\sqrt{10}}{32}=\frac{8(3+\sqrt{10})}{32}=\frac{(3+\sqrt{10})}{4} \\ \beta =\frac{-b-\sqrt{D}}{2a}=\frac{-(-24)-\sqrt{640}}{2\times 16}=\frac{24-8\sqrt{10}}{32}=\frac{8(3-\sqrt{10})}{32}=\frac{(3-\sqrt{10})}{4} \\ \mathrm{Thus}\text{, the roots of the equation are}\frac{(3+\sqrt{10})}{4}\text{and}\frac{(3-\sqrt{10})}{4}\text{.} \end{gathered}$$