Question

find the zeroes of the polynomials and verify the relationship between the zeroes and coefficients 4x^2 - 4x +1​

Answer 1

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Answers

DaInvincible12

DaInvincible12 Ace

Solution

Given

Let the given polynomial be r(x) = 4x² - 4x + 1

Now,

\sf \: f(x) = 4 {x}^{2} - 4x + 1 \\ \\ \implies \: \sf \: f(x) = 4x {}^{2} - 2x - 2x + 1 \\ \\ \implies \: \sf \: f(x) = 2x(2x - 1) - 1(2x - 1) \\ \\ \implies \: \sf \: \: f(x) = (2x - 1)(2x - 1)

From Remainder Theorem,

\sf \: f(x) = 0 \\ \\ \longrightarrow \: \sf \: (2x - 1)(2x - 1) = 0 \\ \\ \longrightarrow \: \boxed { \boxed {\sf \: x = \dfrac{1}{2} }}

Let \sf \alpha \ and \beta are the zeros of the f(x)

\sf \alpha = \dfrac{1}{2}

\sf \: \beta = \dfrac{1}{2}

Now,

Sum of Zeros

\sf \: \alpha + \beta \\ \\ \dashrightarrow \: \sf \: \dfrac{1}{2} + \dfrac{1}{2} \\ \\ \dashrightarrow \: \sf \: 1 \: \sim \: - ( - \dfrac{4}{4} )

Product Of Zeros

\sf \: \alpha \beta \\ \\ \dashrightarrow \sf \: \dfrac{1}{2} \times \dfrac{1}{2} \\ \\ \dashrightarrow \: \sf \: \dfrac{1}{4}

Step-by-step explanation:

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Answer 2

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