If the roots of \[a{{x}^{2}}+bx+c=0\] are \[\alpha ,\beta \] and the roots of \[A{{x}^{2}}+Bx+C=0\]are \[\alpha -k,\beta -k,\]then \[\frac{{{B}^{2}}-4AC}{{{b}^{2}}-4ac}\] is equal to [RPET 1999]

\[{{\left( \frac{A}{a} \right)}^{2}}\]

\[{{\left( \frac{a}{A} \right)}^{2}}\]

If the roots of \[a{{x}^{2}}+bx+c=0\] are \[\alpha ,\beta \] and the r

1.	If the roots of \[a{{x}^{2}}+bx+c=0\] are \[\alpha ,\beta \] and the roots of \[A{{x}^{2}}+Bx+C=0\]are \[\alpha -k,\beta -k,\]then \[\frac{{{B}^{2}}-4AC}{{{b}^{2}}-4ac}\] is equal to [RPET 1999]
A.	0
B.	1
C.	\[{{\left( \frac{A}{a} \right)}^{2}}\]
D.	\[{{\left( \frac{a}{A} \right)}^{2}}\]
Answer» C. \[{{\left( \frac{A}{a} \right)}^{2}}\]

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If the roots of \[a{{x}^{2}}+bx+c=0\] are \[\alpha ,\beta \] and the roots of \[A{{x}^{2}}+Bx+C=0\]are \[\alpha -k,\beta -k,\]then \[\frac{{{B}^{2}}-4AC}{{{b}^{2}}-4ac}\] is equal to [RPET 1999]

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