Question

Find the zeroes of the following quadritic polynomial and verify the relationship between the zeroes and the coefficient: 3x square - x - 4

Answer 1

Question

Find the zeroes of the following quadritic polynomial and verify the relationship between the zeroes and the coefficient: 3x ² - x - 4

Solution

Given:-

• Equation, 3x² - x - 4 = 0

Find:-

• the relationship between the zeroes and the coefficient

Explanation

Equation,

-----> 3x² - x - 4 = 0

-----> 3x² - 4x + 3x - 4 = 0

-----> 3x(x+1)-4(x+1) = 0

-----> (3x-4)(x+1) = 0

-----> (3x -4) = 0. Or, (x+1) = 0

-----> x = 4/3. Or, x = -1

Zeroes of this equation be 4/3 and -1

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the relationship between the zeroes and the coefficient

☛ Sum of zeroes = -B/A

☛ Product of zeroes = C/ A

where,

• A = 3
• B = -1
• C = -4

•°• Sum of zeroes = -(-1)/3

•°• Sum of zeroes = 1/3

➤▸ (4/3) - 1 = 1/3

➤▸ (4-3)/3 = 1/3

➤▸ 1/3 = 1/3

L.H.S. = R.H.S.

And,

•°• Product of zeroes = (-4)/3

•°• Product of zeroes = -4/3

➤▸ (4/3) * (-1) = -4/3

➤▸ -4/3 = -4/3

L.H.S. = R.H.S.

That's proved.

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Answer 2

Step-by-step explanation:

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$\bf \huge \: Question \:$

Find the zeroes of the following quadritic polynomial and verify the relationship between the zeroes and the coefficient: 3x square - x - 4

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$\bf \huge \: given$

• Quadritic polynomial 3x square - x - 4

.

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$\bf \huge \: to \: Find \: \: \: \:$

• Find the zeroes of the following Quadritic polynomial
• verify the relationship between the zeroes and the coefficient: 3x square -x - 4

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$\bf \huge \: Solution \:$

First we will find the zeroes of the quadratic polynomial.

We will use "Split the middle terms":

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$\bf \: {3x}^{2} - x - 4 = 0 \: \\ \bf according \: to \: the \: quadratic \: \\ \bf {3x}^{2} - 4x + 3x - 4 = 0\\ \bf \: x(3x - 4) + 1(3x - 4) = 0\\ \bf \: (x + 1)(3x - 4) = 0 \\ \\ \bf \: \red{\alpha =\frac{4}{3},\beta =--1 \: } \:$

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Now,

Let,

$\bf \: \red{\alpha =\frac{4}{3},\beta =--1 \: }$

Now,

we will verify the relationship between the zeroes and coefficient.

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$\bf\begin{lgathered}\alpha +\beta =\frac{4}{3}-1=\frac{1}{3}\\\\\ \bf \: alpha \beta =-1\times \frac{4}{3}=\frac{-4}{3}\\ \bf \: and\\\\\ \bf \: alpha +\beta =\frac{-b}{a}=\frac{1}{3},\alpha\beta =\frac{c}{a}=\frac{-4}{3}\end{lgathered} </p><p></p><p> \:$

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HENCE PROVED