 Theory of equations
Question

# Given that  are roots of the equation ${\mathrm{Ax}}^{2}-4\mathrm{x}+1=0$, and  the roots of the equation of ${\mathrm{Bx}}^{2}-6\mathrm{x}+1=0$, such that  are in H.P., then

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Solution

## Since $\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma },\mathrm{\delta }$ are in H.P., $1/\mathrm{\alpha },1/\mathrm{\beta },1/\mathrm{\gamma },1/\mathrm{\delta }$ are in A.P. and they maybe taken as $\mathrm{a}-3\mathrm{d},\mathrm{a}-\mathrm{d},\mathrm{a}+\mathrm{d},\mathrm{a}+3\mathrm{d}$. Replacing x by 1/x, we get the equation whose roots are $1/\mathrm{\alpha },1/\mathrm{\beta },1/\mathrm{\gamma },1/\mathrm{\delta }$. Therefore, equation ${\mathrm{x}}^{2}-4\mathrm{x}+\mathrm{A}=0$ has roots a - 3d, a + d and equation ${\mathrm{x}}^{2}-6\mathrm{x}+\mathrm{B}=0$ has roots a - d, a + 3d. Sum of the roots isProduct of the roots is$\begin{array}{r}\left(\mathrm{a}-3\mathrm{d}\right)\left(\mathrm{a}+\mathrm{d}\right)=\mathrm{A}=3\\ \left(\mathrm{a}-\mathrm{d}\right)\left(\mathrm{a}+3\mathrm{d}\right)=\mathrm{B}=8\end{array}$

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