Qudratic Equation-

$Quadratic\quad equation\Longrightarrow a{ x }^{ 2 }+bx+c=0\\ Discriminant\quad D={ b }^{ 2 }-4ac\\ Let\quad its\quad roots\quad be\quad { x }_{ 1 }\quad and\quad { x }_{ 2 }\quad \\ then\quad { x }_{ 1 }+{ x }_{ 2 }=-\frac { b }{ a } \quad ;\quad { x }_{ 1 }{ x }_{ 2 }=\frac { c }{ a } \\ { x }_{ 1 }-{ x }_{ 2 }=m\\ { \left( { x }_{ 1 }-{ { x }_{ 2 } } \right) }^{ 2 }\quad ={ \left( { x }_{ 1 }+{ { x }_{ 2 } } \right) }^{ 2 }-4{ x }_{ 1 }{ x }_{ 2 }\\ \quad \quad \quad \quad { m }^{ 2 }=\frac { { b }^{ 2 } }{ { a }^{ 2 } } -4\frac { c }{ a } \\ \quad \qquad \quad { (ma) }^{ 2 }={ b }^{ 2 }-4ac$

$Roots\quad of\quad a{ x }^{ 2 }+bx+c=0,\quad by\quad quadratic\quad formula\\ =\quad \frac { -b\pm \sqrt { D } }{ 2a } ,\quad where\quad D\quad is\quad the\quad discriminant.\\ Acc.\quad to\quad question,\quad roots\quad differ\quad by\quad 'm'\quad i.e.\\ \frac { -b+\sqrt { D } }{ 2a } \quad -\quad m\quad =\quad \frac { -b-\sqrt { D } }{ 2a } \\ \frac { -b+\sqrt { D } -2am }{ 2a } \quad =\quad \frac { -b-\sqrt { D } }{ 2a } \\ Cancelling\quad '-b'\quad and\quad '2a',\quad we\quad are\quad left\quad with:\\ \sqrt { D } -2am\quad =\quad -\sqrt { D } \\ 2\sqrt { D } \quad =\quad 2am\\ Cancelling\quad '2'\quad and\quad squaring:\\ D={ (am) }^{ 2 }$

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