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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

Thenumberofrealrootsofthequadraticequation(x−4)2+(x−5)2+(x−6)2=0is

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

4,5,6sumof3squaresare0∴eachofthem0∴(x−4)2=0⇒x=4(x−5)2=0⇒x=5(x−6)2=0⇒x=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!

thereisnorealroot!lett=x−5(x−4)2+(x−5)2+(x−6)2=(t+1)2+t2+(t−1)2=3t2+2⩾2i.e.(x−4)2+(x−5)2+(x−6)2=0hasnorealroot!

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

letx−6=t−1,x−5=t,x−4=t+1⇔t2+(t+1)2+(t−1)2=0t2+t2+t2+2=03t2=−2⇔t∉R,sincet2⩾0fort∈R

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