A) Repeated root is 2
B) \[a=10,\text{ }b=-\,20\]
C) \[a=0,\text{ }b=-\,20\]
D) none of these
Correct Answer: C
Solution :
| Let \[f(x)={{x}^{4}}+a{{x}^{3}}-13{{x}^{2}}+bx-4=0\] If \[2+\sqrt{5}\] is one root then other has to be \[2-\sqrt{5}\] Let \[\pm \] be repeated root, then Product of the roots is |
| \[(2+\sqrt{5})(2-\sqrt{5}){{\alpha }^{2}}=-4\]\[\Rightarrow \]\[{{\alpha }^{2}}=4\operatorname{or}\,\alpha =\pm 2\] ...(i) |
| But α can take only one value, so now consider \[\sum{\alpha \beta =-13}\] |
| \[\therefore -1+\left( 4-2\sqrt{5} \right)\alpha +\left( 4+2\sqrt{5} \right)\alpha +{{\alpha }^{2}}=-13\] |
| \[{{\alpha }^{2}}+8\alpha +12=0\] |
| \[\left( \alpha +6 \right)\left( \alpha +2 \right)=0\] |
| \[\alpha =-2\,\,or\,\,-6\] ...(ii) |
| From result (i) and (ii), \[\alpha =-2\] Sum of the roots is, \[2+\sqrt{5}+2-\sqrt{5}+\left( 2 \right)+\left( -2 \right)=\operatorname{a}=0\] Similarly \[b=-20\] |
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