A) p = 1, q = - 56
B) p = - 1, q = - 56
C) p = 1, q = 56
D) p = - 1, q = 56
Correct Answer: B
Solution :
Since roots of the equation \[{{x}^{2}}-5x+16=0\]are \[\alpha ,\beta \]. \[\Rightarrow \alpha +\beta =5\]and \[\alpha \beta =16\] and \[{{\alpha }^{2}}+{{\beta }^{2}}+\frac{\alpha \beta }{2}=-p\] \[\Rightarrow {{(\alpha +\beta )}^{2}}-2\alpha \beta +\frac{\alpha \beta }{2}=-p\]\[\Rightarrow 25-32+8=\,-p\] \[\Rightarrow p=-1\] \[\text{and}\,\,\text{(}{{\alpha }^{\text{2}}}+{{\beta }^{2}})\,\left( \frac{\alpha \beta }{2} \right)=q\] \[\Rightarrow \left[ \,{{(\alpha +\beta )}^{2}}-2\alpha \beta \right]\,\left[ \frac{\alpha \beta }{2} \right]=q\] Þ\[q=[25-32]\frac{16}{2}=-56\] So, \[p=-1,q=-56\].
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