Question

can anyone solve thi. It will be a great help​

Answer 1

Question:

$\sf{Find \ the \ value \ of \ a, \ if \ whenever \ \alpha} \\ \\ \sf{ is \ a \ root \ of \ x^{2} + ax - 1 = 0, \ then \ (\alpha^{2}-2)} \\ \\ \sf{ will \ also \ be \ root \ of \ the \ this \ equation.}$

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Answer:

$\sf{The \ value \ of \ a \ is \ 0.}$

Given:

$\sf{The \ given \ quadratic \ equation \ is} \\ \\ \sf{x^{2}+ax-1=0}$

To find:

$\sf{The \ value \ of \ a \ if \ roots \ of \ the \ equation} \\ \\ \sf{are \ \alpha \ and \ (\alpha^{2}-2).}$

Solution:

$\sf{The \ given \ quadratic \ equation \ is} \\ \\ \sf{x^{2}+ax-1=0} \\ \\ \sf{Roots \ of \ equation \ are \ \alpha \ and \ \alpha^{2}-2} \\ \\ \sf{Here, \ a=1, \ b=a \ and \ c=-1} \\ \\ \boxed{\sf{Product \ of \ roots=\dfrac{c}{a}}} \\ \\ \sf{\therefore{\alpha(\alpha^{2}-2)=-1}} \\ \\ \sf{\therefore{\alpha^{3}-2\alpha+1=0}} \\ \\ \sf{\therefore{\alpha^{3}-\alpha-\alpha+1=0}} \\ \\ \sf{\therefore{\alpha^{2}(\alpha-1)-(\alpha-1)=0}} \\ \\ \sf{\therefore{(\alpha-1)(\alpha^{2}-1)=0}} \\ \\ \sf{\therefore{\alpha=1 \ or \ \alpha=\pm1}} \\ \\ \sf{\therefore{The \ value \ of \ \alpha \ is \ \pm1.}}$

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$\sf{When \ \alpha=1,} \\ \\ \sf{\leadsto{Roots, \ \alpha=1 \ and \alpha^{2}-2=-1}} \\ \\ \boxed{\sf{Sum \ of \ roots=\dfrac{-b}{a}}} \\ \\ \sf{\therefore{(\alpha)+(\alpha^{2}-1)=-a}} \\ \\ \sf{\therefore{1-1=-a}} \\ \\ \sf{\therefore{a=0}} \\ \\ \sf{Verification \ for \ a=0} \\ \\ \sf{\leadsto{x^{2}+0x-1=0}} \\ \\ \sf{x=\pm1} \\ \\ \sf{Since, \ the \ value \ of \ \alpha \ and \ (\alpha^{2}-2)} \\ \sf{is \ 1 \ and \ -1 \ respectively.} \\ \\ \sf{It \ is \ verified \ value \ of \ a \ is \ 0 \ whenever} \\ \sf{roots \ are \ \alpha \ and \ (\alpha^{2}-1).}$

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$\sf{When \ \alpha=-1,} \\ \\ \sf{\leadsto{Roots, \ \alpha=-1 \ and \alpha^{2}-2=-1}} \\ \\ \boxed{\sf{Sum \ of \ roots=\dfrac{-b}{a}}} \\ \\ \sf{\therefore{(\alpha)+(\alpha^{2}-1)=-a}} \\ \\ \sf{\therefore{-1-1=-a}} \\ \\ \sf{\therefore{a=2}} \\ \\ \sf{Verification \ for \ a=2} \\ \\ \sf{\leadsto{x^{2}+2x-1=0}} \\ \\ \sf{\therefore{x=\dfrac{2\pm\sqrt{8}}{2}}} \\ \\ \sf{\therefore{x=1\pm\sqrt2}} \\ \\ \sf{But, \ it's \ not \ satisfying \ values} \\ \sf{ of \ \alpha \ and \ (\alpha^{2}-1).} \\ \\ \sf{\therefore{a\neq 2}}$

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$\purple{\tt{Hence, \ the \ value \ of \ a \ is \ 0.}}$