If the roots of the equation ${ x }^{ 3 }-p{ x }^{ 2 }+qx-r=0$ are in AP, then
- $2{ p }^{ 3 }=9pq-27r$
- $2{ q }^{ 3 }=9pq-27r$
- ${ p }^{ 3 }=9pq-27r$
- $2{ p }^{ 3 }=9pq+27r$
Reveal answer
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A
Correct answer
Explanation
If roots a-d, a, a+d are in AP, their sum is 3a = p, so a = p/3. Substituting a into the equation p(p/3)^2 - q(p/3) + r = 0 leads to the relation 2p^3 = 9pq - 27r.
AI explanation
Let the roots of the cubic equation be a minus d, a, and a plus d. By the sum of roots, p equals 3a, so a equals p divided by 3. The condition for roots in arithmetic progression is that substituting x equals p divided by 3 must satisfy the cubic equation. Substituting p divided by 3 into x cubed minus p x squared plus q x minus r equals 0 and multiplying by 27 yields the relation 2 p cubed equals 9 p q minus 27 r.