Question

If (alpha) and (beta) are zeroes of the polynomial f(x) = ax² + bx + c and (K + alpha) and (K + beta) are zeroes of the polynomial p(x) = Ax² + Bx + C then,

Prove : [ (B² - 4AC)÷(b² - 4ac) = A²/a²

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Answer 1

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Answer 2

Given Quadratic polynomial is ax^2 + bx + c.

Let a, b be the zeroes of the given polynomial.

= > We know that Sum of zeroes = -b/a

        a + b = -b/a

= > We know that Product of zeroes = c/a

       ab = c/a

Now,

We know that By algebraic identity (a - b)^2 = (a + b)^2 - 4ab

 (a - b)^2 = (-b/a)^2 - 4(c/a)

= (a - b)^2  = b²/a² - 4c/a

= (a - b)^2  = b²/a² - 4c/a

= (a - b)^2  = (b²-4ac)/4a²