Question

If the roots of the equation $a{x}^2+2bx+c=0$ be alpha and beta and those of the equation $A{x}^2+2Bx+C=0$ be alpha+delta and beta+delta prove that,
$\frac{b^2-ac}{a^2}=\frac{B^2-AC}{A^2}$

Answer 1

Dear Student,

$$\begin{gathered} {\mathrm{ax}}^2+2\mathrm{bx}+\mathrm{c}=0\mathrm{has}\mathrm{roots}\mathrm{\alpha }\mathrm{and}\mathrm{\beta } \\ \left.\begin{array}{r}\Rightarrow \mathrm{\alpha }+\mathrm{\beta }=-\frac{2\mathrm{b}}{\mathrm{a}}\Rightarrow \frac{\mathrm{b}}{\mathrm{a}}=-\frac{\mathrm{\alpha }+\mathrm{\beta }}{2} \\ \\ \mathrm{and}\mathrm{\alpha \beta }=\frac{\mathrm{c}}{\mathrm{a}}\end{array}\right\}\mathrm{eq}\left(1\right) \\ {\mathrm{Ax}}^2+2\mathrm{Bx}+\mathrm{C}=0\mathrm{has}\mathrm{roots}\mathrm{\alpha }+\mathrm{\delta }\mathrm{and}\mathrm{\beta }+\mathrm{\delta } \\ \left.\begin{array}{r}\Rightarrow \mathrm{\alpha }+\mathrm{\delta }+\mathrm{\beta }+\mathrm{\delta }=-\frac{2\mathrm{B}}{\mathrm{A}}\Rightarrow \frac{\mathrm{B}}{\mathrm{A}}=-\frac{\mathrm{\alpha }+2\mathrm{\delta }+\mathrm{\beta }}{2}\\ \mathrm{and}\left(\mathrm{\alpha }+\mathrm{\delta }\right)\left(\mathrm{\beta }+\mathrm{\delta }\right)=\frac{\mathrm{C}}{\mathrm{A}}\end{array}\right\}\mathrm{eq}\left(2\right) \\ \mathrm{now}\mathrm{To}\mathrm{prove}\frac{{\mathrm{b}}^2-\mathrm{ac}}{{\mathrm{a}}^2}=\frac{{\mathrm{B}}^2-\mathrm{AC}}{{\mathrm{A}}^2} \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{{\mathrm{b}}^2-\mathrm{ac}}{{\mathrm{a}}^2} \\ =\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}-\frac{\mathrm{c}}{\mathrm{a}} \\ ={\left(-\frac{\mathrm{\alpha }+\mathrm{\beta }}{2}\right)}^2-\mathrm{\alpha \beta }\left[\mathrm{using}\mathrm{eq}\left(1\right)\right] \\ ={\left(\frac{\mathrm{\alpha }+\mathrm{\beta }}{2}\right)}^2-\mathrm{\alpha \beta } \\ =\frac{{\mathrm{\alpha }}^2+{\mathrm{\beta }}^2+2\mathrm{\alpha \beta }-4\mathrm{\alpha \beta }}{4} \\ =\frac{{\mathrm{\alpha }}^2+{\mathrm{\beta }}^2-2\mathrm{\alpha \beta }}{4} \\ =\frac{{\left(\mathrm{\alpha }-\mathrm{\beta }\right)}^2}{4} \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{{\mathrm{B}}^2-\mathrm{AC}}{{\mathrm{A}}^2} \\ =\frac{{\mathrm{B}}^2}{{\mathrm{A}}^2}-\frac{\mathrm{C}}{\mathrm{A}} \\ ={\left(-\frac{\mathrm{\alpha }+2\mathrm{\delta }+\mathrm{\beta }}{2}\right)}^2-\left(\mathrm{\alpha }+\mathrm{\delta }\right)\left(\mathrm{\beta }+\mathrm{\delta }\right) \\ ={\left(\frac{\mathrm{\alpha }+2\mathrm{\delta }+\mathrm{\beta }}{2}\right)}^2-\left(\mathrm{\alpha }+\mathrm{\delta }\right)\left(\mathrm{\beta }+\mathrm{\delta }\right) \\ =\frac{{\mathrm{\alpha }}^2+4{\mathrm{\delta }}^2+{\mathrm{\beta }}^2+4\mathrm{\alpha \delta }+4\mathrm{\delta \beta }+2\mathrm{\alpha \beta }}{4}-\left(\mathrm{\alpha \beta }+\mathrm{\alpha \delta }+\mathrm{\delta \beta }+{\mathrm{\delta }}^2\right) \\ =\frac{{\mathrm{\alpha }}^2+4{\mathrm{\delta }}^2+{\mathrm{\beta }}^2+4\mathrm{\alpha \delta }+4\mathrm{\delta \beta }+2\mathrm{\alpha \beta }-4\left(\mathrm{\alpha \beta }+\mathrm{\alpha \delta }+\mathrm{\delta \beta }+{\mathrm{\delta }}^2\right)}{4} \\ =\frac{{\mathrm{\alpha }}^2+4{\mathrm{\delta }}^2+{\mathrm{\beta }}^2+4\mathrm{\alpha \delta }+4\mathrm{\delta \beta }+2\mathrm{\alpha \beta }-4\mathrm{\alpha \beta }-4\mathrm{\alpha \delta }-4\mathrm{\delta \beta }-4{\mathrm{\delta }}^2}{4} \\ =\frac{{\mathrm{\alpha }}^2+{\mathrm{\beta }}^2+2\mathrm{\alpha \beta }-4\mathrm{\alpha \beta }}{4} \\ =\frac{{\mathrm{\alpha }}^2+{\mathrm{\beta }}^2-2\mathrm{\alpha \beta }}{4} \\ =\frac{{\left(\mathrm{\alpha }-\mathrm{\beta }\right)}^2}{4} \\ \Rightarrow \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}. \end{gathered}$$

Regards