Is it true that if a, b, c are in arithmetic progression, and if the polynomials (b−c)x2+(c−a)x+(a−b) and 2(c+a)x2+(b+c)x have a common root, then a2, c2, b2 must be un arithmetic progression also?

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Is it true that if a, b, c are in arithmetic progression, and if the polynomials (b−c)x2+(c−a)x+(a−b) and 2(c+a)x2+(b+c)x have a common root, then a2, c2, b2 must be un arithmetic progression also?

Anderson Higgs · Accepted Answer

Step 1This one’s a damn easy.Let’s sail in the seas of AP.It is stated that a,b,c are in AP. So2b=a+cNow let’s factorize the first quadratic polynomial with a bit of skill:(b−c)x2+(c−a)x+(a−b)=0(b−c)x2−(b−c)x−(a−b)x+(a−b)=0(b−c)x(x−1)−(a−b)(x−1)=0((b−c)x−(a−b))(x−1)=0x=b−ca−b, 1x=a−ba−b,x=1, 1Now, let&#039;s do something with the second one:2(c+a)x2+(b+c)x=0(x)(2(c+a)x+(b+c))=0x=0, −b+c2(c+a)x=0, −b+c2×2bx=0, −b+c4bNow it is completely sure that:−b+c4b=1−4b=b+cc=−5basa+c=2ba−5b=2ba=7bSo,c2−a2=25b2−49b2=−24b2b2−c2=b2−25b2=−24b2So,c2−a2=b2−c2So, it is sure that a2, c2, b2 are in AP.

zahrkao8vm · Accepted Answer

Step 1a, b and c in arithmetic progression means:a=b−rtherefore b−a=rc=b+rtherefore c−b=rand c−a=2rand c+a=2bThen the first polynomial can me reduced:(b−c)x2+(c−a)x+(a−b)=−rx2+2rx−rThe roots for−rx2+2rx−r=0are coincident and x=1The second polynomial:2(c+a)x2+(b+c)x=4bx2+(2b+r)xAnd the root for4bx2+(2b+r)x=0arex=0andx=−(2b+r)(4b) with b≠0Since they have one common root, that would be x=1Ergo:1=−2b+r4br=−6bSincea=b−r then a=7bandc=b+r then c=−5bAnd the progression a2, b2, c2 is 25b2, b2, 49b2So the difference between the second and the first terms is 24b2, as is the difference between the first and the third terms.Then, they can form an Arithmetic Progression, but they are not in order, so they are not in Arithmetic Progression.

Is it true that if a, b, c are in

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