Question

Find the zeros of the following quadratic polynomial and verify the relationship between the zeros and the cooperation 3 x square minus x minus 4

Answer 1

Answer:

Given :-

• 3x² - x - 4

To Find :-

• What is the zeros of the quadratic polynomial and verify the relationship between the zeroes and co-efficient.

Solution :-

Given Equation :

$\bigstar\: \: \bf{3x^2 - x - 4}$

Let,

$\implies\sf<strong> </strong>p(x) =\: 3x^2 - x - 4$

Zero of the polynomial is the value of x, where p(x) = 0

By putting p(x) = 0, we get :

$\implies \bf{3x^2 - x - 4 =\: 0}$

$\implies \sf 3x^2 - (4 - 3)x - 4 =\: 0$

$\implies \sf 3x^2 - 4x + 3x - 4 =\: 0\: \: \bigg\lgroup \small\bold{By\: splitting\: middle\: term}\bigg\rgroup$

$\implies \sf x(3x - 4) + 1(3x - 4) =\: 0$

$\implies \sf (3x - 4)(x + 1) =\: 0$

$\longrightarrow \bf 3x - 4 =\: 0$

$\longrightarrow \sf 3x =\: 4$

$\longrightarrow \sf \bold{\red{x =\: \dfrac{4}{3}}}$

$\longrightarrow \bf x + 1 =\: 0$

$\longrightarrow \sf \bold{\red{x =\: - 1}}$

${\small{\bold{\underline{\therefore\: The\: zeroes\: of\: quadratic\: polynomial\: is\: \dfrac{4}{3}\: and\: - 1\: respectively\: .}}}}$

Hence, we get :

• α = 4/3
• β = - 1

Now, we have to verify the relationship between the zeroes and co-efficient :

Given Equation :

$\bigstar\: \: \bf{3x^2 - x - 4}$

By comparing with ax² + bx + c we get,

• a = 3
• b = - 1
• c = - 4

❒ Sum Of Zeroes :

$\clubsuit$ Sum of Zeroes Formula :

$\mapsto \sf\boxed{\bold{\pink{Sum\: of\: Zeroes\: (\alpha + \beta) =\: \dfrac{- b}{a}}}}$

We have :

• a = 3
• b = - 1

According to the question by using the formula we get,

$\longrightarrow \sf \dfrac{4}{3} + (- 1) =\: \dfrac{- (- 1)}{3}$

$\longrightarrow \sf \dfrac{4}{3} - 1 =\: \dfrac{1}{3}$

$\longrightarrow \sf \dfrac{4 - 3}{3} =\: \dfrac{1}{3}$

$\longrightarrow \sf\bold{\purple{\dfrac{1}{3} =\: \dfrac{1}{3}}}$

Hence, Verified.

❒ Product Of Zeroes :

$\clubsuit$ Product Of Zeroes Formula :

$\mapsto \sf\boxed{\bold{\pink{Product\: Of\: Zeroes\: (\alpha\beta) =\: \dfrac{c}{a}}}}$

We have :

• a = 3
• c = - 4

According to the question by using the formula we get,

$\longrightarrow \sf \dfrac{4}{3} \times (- 1) =\: \dfrac{- 4}{3}$

$\longrightarrow \sf\bold{\purple{ \dfrac{- 4}{3} =\: \dfrac{- 4}{3}}}$

Hence, Verified.

Answer 2

Step-by-step explanation:

$let \: f (x) = 3x {}^{2} - x - 4$

$= x(3x - 4) + 1(3x - 4)$

$= (3x - 4)(x + 1)$

$to \: find \: the \: zeroes, Let f(x) = 0$

$(3x - 4) = 0 \: or \: (x + 1) = 0$

$\huge\underline\mathtt\red{x = \frac{4}{3} \: or \: x = - 1}$

So the zeros of f(x) are 4/3 or -1

Again,

Sum of zeros=

$\frac{4}{3} + ( - 1) = \frac{1}{3} = \frac{ - b}{a}$

Product of zeros=

$\frac{4}{3} \times ( - 1) = \frac{ - 4}{3} = \frac{c}{a}$