Given that α, γ are roots of the equation $$Ax2$$−$$4x+1=0$$, and β, δ the roots of the equation of $$Bx2$$−$$6x+1=0$$, such that α,β,γ, and δ are in H.P., then

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Given that α, γ are roots of the equation $$Ax2$$−$$4x+1=0$$, and β, δ the roots of the equation of $$Bx2$$−$$6x+1=0$$, such that α,β,γ, and δ are in H.P., then

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Since α,β,γ,δ are in H.P., $$1/$$α,$$1/$$β,$$1/$$γ,$$1/$$δ are in A.P. and they maybe taken as a−$$3d$$,a−d,$$a+d$$,$$a+3d$$. Replacing x by $$1/x$$, we get the equation whose roots are $$1/$$α,$$1/$$β,$$1/$$γ,$$1/$$δ. Therefore, equation $$x2$$−$$4x+A=0$$ has roots a $$-$$ $$3d$$, a $$+$$ d and equation $$x2$$−$$6x+B=0$$ has roots a $$-$$ d, a $$+$$ $$3d$$. Sum of the roots is    $$2(a$$−$$d)=4$$,$$2(a+d)=6$$∴     $$a=5/2$$,$$d=1/2Product$$ of the roots $$is(a$$−$$3d)(a+d)=A=3(a$$−$$d)(a+3d)=B=8$$

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

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Given that α, γ are roots of the equation Ax2−4x+1=0, and β, δ the roots of the equation of Bx2−6x+1=0, such that α,β,γ, and δ are in H.P., then

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