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The-number-of-real-roots-of-the-quadratic-equation-x-4-2-x-5-2-x-6-2-0-is-

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Question Number 98056 by PengagumRahasiamu last updated on 11/Jun/20
The number of real roots of the quadratic equation (x−4)^2 +(x−5)^2 +(x−6)^2 =0 is
$$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{real}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{quadratic} \\ $$$$\mathrm{equation}\:\left({x}−\mathrm{4}\right)^{\mathrm{2}} +\left({x}−\mathrm{5}\right)^{\mathrm{2}} +\left({x}−\mathrm{6}\right)^{\mathrm{2}} =\mathrm{0}\:\mathrm{is} \\ $$
Commented by som(math1967) last updated on 11/Jun/20
4,5,6 sum of 3 squares are 0 ∴each of them 0 ∴(x−4)^2 =0⇒x=4 (x−5)^2 =0⇒x=5 (x−6)^2 =0 ⇒x=6
$$\mathrm{4},\mathrm{5},\mathrm{6} \\ $$$$\mathrm{sum}\:\mathrm{of}\:\mathrm{3}\:\mathrm{squares}\:\mathrm{are}\:\mathrm{0} \\ $$$$\therefore\mathrm{each}\:\mathrm{of}\:\mathrm{them}\:\mathrm{0} \\ $$$$\therefore\left(\mathrm{x}−\mathrm{4}\right)^{\mathrm{2}} =\mathrm{0}\Rightarrow\mathrm{x}=\mathrm{4} \\ $$$$\left(\mathrm{x}−\mathrm{5}\right)^{\mathrm{2}} =\mathrm{0}\Rightarrow\mathrm{x}=\mathrm{5} \\ $$$$\left(\mathrm{x}−\mathrm{6}\right)^{\mathrm{2}} =\mathrm{0}\:\Rightarrow\mathrm{x}=\mathrm{6} \\ $$
Commented by mr W last updated on 11/Jun/20
there is no real root! let t=x−5 (x−4)^2 +(x−5)^2 +(x−6)^2 =(t+1)^2 +t^2 +(t−1)^2 =3t^2 +2 ≥2 i.e. (x−4)^2 +(x−5)^2 +(x−6)^2 =0 has no real root!
$${there}\:{is}\:{no}\:{real}\:{root}! \\ $$$${let}\:{t}={x}−\mathrm{5} \\ $$$$\left({x}−\mathrm{4}\right)^{\mathrm{2}} +\left({x}−\mathrm{5}\right)^{\mathrm{2}} +\left({x}−\mathrm{6}\right)^{\mathrm{2}} \\ $$$$=\left({t}+\mathrm{1}\right)^{\mathrm{2}} +{t}^{\mathrm{2}} +\left({t}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$=\mathrm{3}{t}^{\mathrm{2}} +\mathrm{2} \\ $$$$\geqslant\mathrm{2} \\ $$$${i}.{e}.\:\left({x}−\mathrm{4}\right)^{\mathrm{2}} +\left({x}−\mathrm{5}\right)^{\mathrm{2}} +\left({x}−\mathrm{6}\right)^{\mathrm{2}} =\mathrm{0}\:{has} \\ $$$${no}\:{real}\:{root}! \\ $$
Commented by bemath last updated on 11/Jun/20
let x−6 = t−1 , x−5 = t , x−4 = t+1 ⇔ t^2 +(t+1)^2 +(t−1)^2 =0 t^2 +t^2 +t^2 +2 = 0 3t^2 = −2 ⇔ t ∉R , since t^2 ≥ 0 for t ∈R
$$\mathrm{let}\:\mathrm{x}−\mathrm{6}\:=\:\mathrm{t}−\mathrm{1}\:,\:\mathrm{x}−\mathrm{5}\:=\:\mathrm{t}\:,\:\mathrm{x}−\mathrm{4}\:=\:\mathrm{t}+\mathrm{1} \\ $$$$\Leftrightarrow\:\mathrm{t}^{\mathrm{2}} +\left(\mathrm{t}+\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{t}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{t}^{\mathrm{2}} +\mathrm{t}^{\mathrm{2}} +\mathrm{t}^{\mathrm{2}} +\mathrm{2}\:=\:\mathrm{0} \\ $$$$\mathrm{3t}^{\mathrm{2}} \:=\:−\mathrm{2}\:\Leftrightarrow\:\mathrm{t}\:\notin\mathbb{R}\:,\:\mathrm{since}\:\mathrm{t}^{\mathrm{2}} \:\geqslant\:\mathrm{0} \\ $$$$\mathrm{for}\:\mathrm{t}\:\in\mathbb{R} \\ $$

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