Question

Q. If $\alpha ,\beta (where\alpha <\beta )arerootsoftheequation{x}^2+bx+c=0where(c<0<b)then$
$$\begin{gathered} 1.0<\alpha <\beta \\ 2.\alpha <0<\beta <\left|\alpha \right| \\ 3.\alpha <\beta <0 \\ 4.\alpha <0<\left|\alpha \right|<\beta \end{gathered}$$

Answer 1

Dear student
$$\begin{gathered} \mathrm{Let}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2+\mathrm{bx}+\mathrm{c}=0 \\ \mathrm{Discriminant},\mathrm{D}={\mathrm{B}}^2-4\mathrm{AC}\mathrm{where}\mathrm{A}=1,\mathrm{B}=\mathrm{b}\mathrm{and}\mathrm{C}=\mathrm{c} \\ \mathrm{So},\mathrm{D}={\mathrm{b}}^2-4\mathrm{ac} \\ ={\mathrm{b}}^2+4\mathrm{ac}\left[\because \mathrm{c}<0\right] \\ >0 \\ \mathrm{So},\mathrm{both}\mathrm{are}\mathrm{real}\mathrm{and}\mathrm{positive} \\ \mathrm{So},\mathrm{\alpha },\mathrm{\beta }>0 \\ \mathrm{Hence}0<\mathrm{\alpha }<\mathrm{\beta }\left[\because \mathrm{\alpha }<\mathrm{\beta }\right] \end{gathered}$$
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