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Question Number 37139 by nishant last updated on 09/Jun/18
if α , β are the roots of the quadratic equation ax^2 +bx+c =0 then find the quadratic equation whose roots are α^(2 ) , β^2
$${if}\:\alpha\:,\:\beta\:\:{are}\:{the}\:{roots}\:{of}\:{the}\:{quadratic} \\ $$$${equation}\:{ax}^{\mathrm{2}} +{bx}+{c}\:=\mathrm{0}\:{then}\:\:{find} \\ $$$${the}\:{quadratic}\:{equation}\:{whose}\:{roots} \\ $$$${are}\:\:\alpha^{\mathrm{2}\:\:\:} ,\:\beta^{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 09/Jun/18
x^2 −x(α^2 +β^2 )+α^2 β^2 =0 x^2 −x{(α+β)^2 −2αβ}+(αβ)^2 =0× x^2 −x{((b^2 /a^2 )−((2c)/a))}+(c^2 /a^2 )=0 a^2 x^2 −x(b^2 −2ac)+c^2 =0
$${x}^{\mathrm{2}} −{x}\left(\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} \right)+\alpha^{\mathrm{2}} \beta^{\mathrm{2}} =\mathrm{0} \\ $$$${x}^{\mathrm{2}} −{x}\left\{\left(\alpha+\beta\right)^{\mathrm{2}} −\mathrm{2}\alpha\beta\right\}+\left(\alpha\beta\right)^{\mathrm{2}} =\mathrm{0}× \\ $$$${x}^{\mathrm{2}} −{x}\left\{\left(\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\frac{\mathrm{2}{c}}{{a}}\right)\right\}+\frac{{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} }=\mathrm{0} \\ $$$${a}^{\mathrm{2}} {x}^{\mathrm{2}} −{x}\left({b}^{\mathrm{2}} −\mathrm{2}{ac}\right)+{c}^{\mathrm{2}} =\mathrm{0} \\ $$
Answered by Joel579 last updated on 09/Jun/18
α + β = −(b/a), αβ = (c/a) The new quadratic equation: x^2 − (α^2 + β^2 )x + α^2 β^2 = 0 x^2 − [(α + β)^2 − 2αβ]x + (αβ)^2 = 0 x^2 − ((b^2 /a^2 ) − ((2c)/a))x + (c^2 /a^2 ) = 0 a^2 x^2 − (b^2 − 2ac)x + c^2 = 0
$$\alpha\:+\:\beta\:=\:−\frac{{b}}{{a}},\:\:\:\:\alpha\beta\:=\:\frac{{c}}{{a}} \\ $$$$ \\ $$$$\mathrm{The}\:\mathrm{new}\:\mathrm{quadratic}\:\mathrm{equation}: \\ $$$${x}^{\mathrm{2}} \:−\:\left(\alpha^{\mathrm{2}} \:+\:\beta^{\mathrm{2}} \right){x}\:+\:\alpha^{\mathrm{2}} \beta^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$${x}^{\mathrm{2}} \:−\:\left[\left(\alpha\:+\:\beta\right)^{\mathrm{2}} \:−\:\mathrm{2}\alpha\beta\right]{x}\:+\:\left(\alpha\beta\right)^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$${x}^{\mathrm{2}} \:−\:\left(\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\:−\:\frac{\mathrm{2}{c}}{{a}}\right){x}\:+\:\frac{{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\:=\:\mathrm{0} \\ $$$${a}^{\mathrm{2}} {x}^{\mathrm{2}} \:−\:\left({b}^{\mathrm{2}} \:−\:\mathrm{2}{ac}\right){x}\:+\:{c}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$

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