If the roots of x³-px²+qx-r = 0 are in A.P, then __________

Correct Answer is : 2p3−9pq+27r2p³-9pq+27r2p3−9pq+27r = 000

Correct Answer is : 2p3−9pq+27r2p³-9pq+27r2p3−9pq+27r = 000

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AP EAMCET Engineering 22 Sep 2020 Shift 2 Paper

Section: Mathematics

Question : 43 of 160

Marks: +1, -0

If the roots of

x^3-px^2+qx-r

0

are in A.P, then __________

Solution:

Let

a-d, a, a+d

are roots of

x^{3}-p x^{2}+q x-r=0

...(i)

Sum of roots

=-\;\frac{b}{a}

a-d+a+a+d=-\;\frac{-p}{1}

3a=p

a=\;\frac{p}{3}

Since,

a=\;\frac{P}{3}

should be satisfied by given equation.

So, put

x=\;\frac{p}{3}

in Eq. (i)

\left(\;\frac{p}{3}\right)^{3}-P\left(\;\frac{p}{3}\right)^{2}+q\left(\;\frac{p}{3}\right)-r=0

\;\frac{p^{3}}{27}-\;\frac{p^{3}}{9}+\;\frac{p q}{3}-r=0

p^{3}-3 p^{3}+9 p q-27 r=0

-2 p^{3}+9 p q-27 r=0

2 p^{3}-9 p q+27 r=0

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