For example if α, β, γ are roots of ax3 + bx2 + cx + d = 0 to find equation whose roots are 1/α, 1/β, 1/γ, we put 1α = y ⇒ α = 1y . As α is a root of ax3 + bx2 + cx + d = 0, we get ay3 + by2 + cy + d = 0 or a + by + cy2 + dy3 = 0 This is the desired equation.The same result holds for all polynomial equations.If α, β are roots of ax2 + bx + c = 0 the roots of a(x − 1)2 + b(x − 1)(x − 2) + c(x − 2)2 = 0 are
Question
For example if α, β, γ are roots of ax3 + bx2 + cx + d = 0 to find equation whose roots are 1/α, 1/β, 1/γ, we put 1α = y ⇒ α = 1y . As α is a root of ax3 + bx2 + cx + d = 0, we get ay3 + by2 + cy + d = 0 or a + by + cy2 + dy3 = 0 This is the desired equation.The same result holds for all polynomial equations.If α, β are roots of ax2 + bx + c = 0 the roots of a(x − 1)2 + b(x − 1)(x − 2) + c(x − 2)2 = 0 are
Solution
The roots of the equation a(x - 1)² + b(x - 1)(x - 2) + c(x - 2)² = 0 are α + 1 and β + 2. This is because the transformation x → x - 1 shifts the roots of the equation ax² + bx + c = 0 by 1 to the right, and the transformation x → x - 2 shifts the roots of the equation ax² + bx + c = 0 by 2 to the right. Therefore, the roots of the transformed equation are the original roots shifted by 1 and 2, respectively.
Similar Questions
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