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If the Zeros of the Polynomial F(X) = Ax3 + 3bx2 + 3cx + D Are in A.P., Prove that 2b3 − 3abc + A2d = 0. - Mathematics

If the zeros of the polynomial f(x) = ax3 + 3bx2 + 3cx + d are in A.P., prove that 2b3 − 3abc + a2d = 0.

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Solution

Let a - d, a and a + d be the zeros of the polynomial f(x). Then,

Sum of the zeroes `=("coefficient of "x^2)/("coefficient of "x^3)`

`a-d+a+a+d=(-3b)/a`

`3a=(-3b)/a`

`a=(-3b)/axx1/3`

`a=(-b)/a`

Since a is a zero of the polynomial f(x).

Therefore,

f(x) = ax3 + 3bx2 + 3cx + d

f(a) = 0

f(a) = aa3 + 3ba2 + 3ca + d

aa3 + 3ba2 + 3ca + d = 0

`a((-b)/a)^3+3bxx((-b)/a)^2+3cxx((-b)/a)+d=0`

`axx(-b)/axx(-b)/axx(-b)/a+3xxbxx(-b)/axx(-b)/a+3xxcxx(-b)/a+d=0`

`(-b^3)/a^2+(3b^3)/a^2-(3cb)/a+d=0`

`(-b^3+3b^3-3abc+a^2d)/a^2=0`

2b3 − 3abc + a2d = 0 xa2

2b3 − 3abc + a2d = 0

Hence, it is proved that 2b3 − 3abc + a2d = 0

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APPEARS IN

RD Sharma Class 10 Maths
Chapter 2 Polynomials
Exercise 2.2 | Q 5 | Page 43
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