Question

If 2p - 9pq + 27r = 0 then prove that the roots of the equations txe - Qx2 + px - 1 = 0 are in
H.P.
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Answer 1

Answer:

If 2p³-9pq+27r= 0, then can you prove that the roots of the equation rx³-qx²+px-1=0 are in H.P.?

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1 Answer



Rhea Saikia, studied Mathematics at University of Delhi (2017)

Answered July 20, 2017 · Author has 147 answers and 315.9K answer views

A2A

If a,b and c are the roots of the equation then we need to show that a, b and c are in HP or 1/a, 1/b and 1/c are in AP.

Let, 1/a=x, 1/b=y and 1/c=z. To show x,y and z are in AP

1/a, 1/b and 1/c are nothing but the roots of r/x3−q/x2+p/x−1=0r/x3−q/x2+p/x−1=0i.ex3−px2+qx−r=0x3−px2+qx−r=0

Now, by Vieta's Formulas we have-

x+y+z=px+y+z=p

xy+yz+zx=qxy+yz+zx=q

xyz=rxyz=r

Now substituting the values of p, q and r in

2p3−9pq+27r=0,2p3−9pq+27r=0,we get-

2(x+y+z)3−9(x+y+z)(xy+yz+zx)+27xyz2(x+y+z)3−9(x+y+z)(xy+yz+zx)+27xyz

⟹(x+y+z)[2(x2+y2+z2)−5(xy+yz+zx)]+27xyz=0⟹(x+y+z)[2(x2+y2+z2)−5(xy+yz+zx)]+27xyz=0

Simplify further and we get-

⟹(x+y−2

Step-by-step explanation:

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