A) \[{{\left( \frac{A}{a} \right)}^{2}}\]
B) \[x\le 0\]
C) \[0\]
D) \[1\]
Correct Answer: A
Solution :
Roots of the equation\[a{{x}^{2}}+bx+c=0\]are\[\alpha ,\beta .\] \[\therefore \] \[\alpha +\beta =-\frac{b}{a}\]and \[\alpha \beta =\frac{c}{a}\] And roots of the equation\[A{{x}^{2}}+Bx+C=0\]are\[\alpha -k,\beta -k\] \[\therefore \] \[(\alpha -k)+(\beta -k)=-\frac{B}{A}\] \[\Rightarrow \] \[(\alpha -k).(\beta -k)=\frac{C}{A}\] Now, \[(\alpha -\beta )={{(\alpha +\beta )}^{2}}-4\alpha \beta \] \[={{\left( -\frac{b}{a} \right)}^{2}}-\frac{4c}{a}\]. \[=\frac{{{b}^{2}}-4ac}{{{a}^{2}}}\] ?..(i) and \[{{(\alpha -\beta )}^{2}}={{\{(\alpha -k)-(\beta -k)\}}^{2}}\] \[={{\{(\alpha -k)+(\beta -k)\}}^{2}}-4(\alpha -k)(\beta -k)\] \[={{\left\{ -\frac{B}{A} \right\}}^{2}}-4\frac{C}{A}=\frac{{{B}^{2}}-4AC}{{{A}^{2}}}\] ?.(ii) From Eqs. (i) and (ii), \[\frac{{{B}^{2}}-4AC}{{{A}^{2}}}=\frac{{{b}^{2}}-4ac}{{{a}^{2}}}\] \[\Rightarrow \] \[\frac{{{B}^{2}}-4AC}{{{b}^{2}}-4ac}=\frac{{{A}^{2}}}{{{a}^{2}}}\] \[\Rightarrow \] \[\frac{{{B}^{2}}-4AC}{{{b}^{2}}-4ac}={{\left( \frac{A}{a} \right)}^{2}}\]
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