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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
Let $\alpha, \beta$ be the roots of the equation
$ x^{2} + 2\sqrt{2}x - 1 = 0 $.
The quadratic equation whose roots are
$\alpha^{4} + \beta^{4}$ and $\dfrac{1}{10} (\alpha^{6} + \beta^{6})$ is:
Solution
Solution:
$ x^2 + 2\sqrt{2}x - 1 = 0 $
$ \alpha + \beta = -2\sqrt{2}, \;\; \alpha\beta = -1 $
$ \alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = 8 + 2 = 10 $
$ \alpha^4 + \beta^4 = (\alpha^2+\beta^2)^2 - 2(\alpha\beta)^2 = 100 - 2 = 98 $
Now,
$ \alpha^3 + \beta^3 = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta) $
$ = (-2\sqrt{2})^3 - 3(-1)(-2\sqrt{2}) = -16\sqrt{2} - 6\sqrt{2} = -22\sqrt{2} $
$ \alpha^6 + \beta^6 = (\alpha^3+\beta^3)^2 - 2(\alpha\beta)^3 $
$ = ( -22\sqrt{2} )^2 - 2(-1)^3 = 968 + 2 = 970 $
Second root:
$ \frac{1}{10}(\alpha^6 + \beta^6) = 97 $
Sum of roots = $ 98 + 97 = 195 $
Product = $ 98 \times 97 = 9506 $
Required equation:
$ x^2 - 195x + 9506 = 0 $
$ \boxed{x^2 - 195x + 9506 = 0} $
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