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Question Number 206795 by BaliramKumar last updated on 25/Apr/24

Commented by BaliramKumar last updated on 25/Apr/24

can be (a) right???

canbe(a)right???

Commented by A5T last updated on 25/Apr/24

x_1 +x_2 =2; x_1 x_2 =−100 ⇒x^2 −2x−100=0

x1+x2=2;x1x2=−100⇒x2−2x−100=0

Commented by A5T last updated on 25/Apr/24

Let the quadratic equation be f(x)=ax^2 +bx+c Then ((−b)/a)=2; (c/a)=−100 ⇒f(x)=a(x^2 −2x−100) x_1 +x_2 =2⇒x_1 =2−x_2 x_1 x_2 =−100⇒(2−x_2 )x_2 =100⇒x_2 ^2 −2x_2 −100=0 ⇒x_2 =1+_− (√(101)) ⇒(x_1 ,x_2 )=(1+(√(101)),1−(√(101))) upto symmetry. The roots will always be the same for all polynomials, so it may contradict (a) which wants it to have different roots.

Letthequadraticequationbef(x)=ax2+bx+cThen−ba=2;ca=−100⇒f(x)=a(x2−2x−100)x1+x2=2⇒x1=2−x2x1x2=−100⇒(2−x2)x2=100⇒x22−2x2−100=0⇒x2=1+−101⇒(x1,x2)=(1+101,1−101)uptosymmetry.Therootswillalwaysbethesameforallpolynomials,soitmaycontradict(a)whichwantsittohavedifferentroots.

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