Question

shows that 1/2-3/2 are the zeroes of the polynomial 4x² + 4x-3 and verify the relationship between zeros and coefficient of the polynomial​

Answer 1

To Show :-

• (1/2) & (-3/2) are the zeroes of the polynomial 4x² + 4x-3 and verify the relationship between zeros and coefficient of the polynomial .. ?

Solution :-

Put The Quadratic Equation to Zero First.

→ 4x² + 4x- 3 = 0

Splitting The Middle Term Now,

→ 4x² + 6x - 2x - 3 = 0

→ 2x(2x + 3) - 1(2x + 3) = 0

→ (2x + 3)(2x - 1) = 0

Putting Both Equal to Zero now,

→ 2x + 3 = 0

→ 2x = (-3)

→ x = (-3/2)

And,

→ (2x - 1) = 0

→ 2x = 1

→ x = (1/2)

Hence, We can say That The roots of The Given Quadratic Equation are (1/2) & (-3/2).

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Now, First Relation is :-

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

Putting both values :-

→ (1/2 + (-3/2) = (-4)/4

→ (1-3)/2 = (-1)

→ (-2/2) = (-1)

→ (-1) = (-1) ✪✪ Hence Verified. ✪✪

Second Relation :-

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

Putting both Values :-

→ (1/2) * (-3/2) = (-3) / 4

→ (-3)/4 = (-3)/4 ✪✪ Hence Verified. ✪✪

_______________________

Answer 2

Answer:

$\underline{\bigstar\:\textsf{Zeroes of the Polynomial :}}$

$:\implies\tt 4x^2+4x-3=0\\\\\\:\implies\tt 4x^2+(6-2)x-3=0\\\\\\:\implies\tt 4x^2+6x-2x-3=0\\\\\\:\implies\tt 2x(2x+3)-1(2x+3)=0\\\\\\:\implies\tt (2x+3)(2x-1)=0\\\\\\:\implies\tt x=\dfrac{-\:3}{2}\quad and\quad\dfrac{1}{2}$

$\rule{160}{1}$

$\underline{\bigstar\:\textsf{Relation b/w zeroes and coefficient :}}$

$\qquad\underline{\bf{\dag}\:\:\textsf{Sum of Zeroes :}}\\\\\dashrightarrow\tt\:\: \alpha+\beta = \dfrac{-\:b}{a}\\\\\\\dashrightarrow\tt\:\: \dfrac{ -\:3}{2} + \dfrac{1}{2}= \dfrac{ - \:4}{4}\\\\\\\dashrightarrow\tt\:\: \dfrac{(- 3 + 1)}{2} = - 1\\\\\\\dashrightarrow\tt\:\: \dfrac{ - 2}{2} = - 1\\\\\\\dashrightarrow\:\:\underline{\boxed{\red{\tt - 1 = - 1}}} \\\\\\\qquad\underline{\bf{\dag}\:\:\textsf{Product of Zeroes :}}\\\\\dashrightarrow\tt\:\: \alpha \times \beta = \dfrac{c}{a}\\\\\\\dashrightarrow\tt\:\: \dfrac{ - 3}{2} \times \dfrac{1}{2} = \dfrac{ - \:3}{4}\\\\\\\dashrightarrow\:\:\underline{\boxed{ \red{\tt \dfrac{ - 3}{4} = \dfrac{ - 3}{4}}}}$