Question

Find the zeroes of the quadratic polynomial 4x²-4x+1 and verify the relation between its zeroes and coefficients

Answer 1

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

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

Answer 2

||✪✪ QUESTION ✪✪||

Find the zeroes of the quadratic polynomial 4x²-4x+1 and verify the relation between its zeroes and coefficients ?

|| ✰✰ ANSWER ✰✰ ||

Given , quadratic polynomial is 4x²-4x+1.

To Find The Number of Zeros Put The polynomial Equals to 0 First.

→ 4x² - 4x + 1 = 0

Now, Splitting The Middle Term , we get,

→ 4x² - 2x - 2x + 1 = 0

→ 2x(2x - 1) - 1(2x - 1) = 0

→ (2x - 1)(2x - 1) = 0

Putting Both Equal to Zero now, we get,

→ 2x - 1 = 0

→ 2x = 1

→ x = (1/2)

Hence, Both Zeros of Quadratic Equation Are (1/2).

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Now, First Relation is :-

→ Sum of Zeros = - (coefficient of x) /(coefficient of x²)

Putting both values ,

→ (1/2 + 1/2) = -(-4)/4

→ 1 = 1 ✪✪ Hence Verified. ✪✪

Second Relation :-

→ Product Of Zeros = Constant Term / (coefficient of x²)

Putting both Values ,

→ (1/2) * (1/2) = (1) / (4)

→ 1/4 = 1/4 ✪✪ Hence Verified. ✪✪

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