A) \[\frac{{{b}^{2}}-2ac}{{{a}^{2}}}\]
B) \[\frac{{{b}^{2}}-2ac}{{{c}^{2}}}\]
C) \[\frac{{{b}^{2}}+2ac}{{{a}^{2}}}\]
D) \[\frac{{{b}^{2}}+2ac}{{{c}^{2}}}\]
Correct Answer: B
Solution :
| [b] Since \[\alpha ,\beta \]are zeroes of the polynomial |
| \[f(x)=a{{x}^{2}}+bx+c\] |
| \[\therefore \] Sum of zeroes \[(\alpha +\beta )=-\frac{b}{a}\] ...(1) |
| and product of zeroes \[(\alpha \beta )=\frac{c}{a}\] ...(2) |
| Now, \[\frac{1}{{{\alpha }^{2}}}+\frac{1}{{{\beta }^{2}}}=\frac{{{\beta }^{2}}+{{\alpha }^{2}}}{{{\alpha }^{2}}{{\beta }^{2}}}=\frac{{{(\alpha +\beta )}^{2}}-2\alpha \beta }{{{(\alpha \beta )}^{2}}}\] |
| \[=\frac{{{\left( -\frac{b}{a} \right)}^{2}}-2\frac{c}{a}}{{{\left( \frac{c}{a} \right)}^{2}}}\] [using (1) and (2)] |
| \[=\frac{\frac{{{b}^{2}}}{{{a}^{2}}}-\frac{2c}{a}}{\frac{{{c}^{2}}}{{{a}^{2}}}}=\frac{\frac{{{b}^{2}}-2ac}{{{a}^{2}}}}{\frac{{{c}^{2}}}{{{a}^{2}}}}=\frac{{{b}^{2}}-2ac}{{{c}^{2}}}\] |
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