Question

find the zero of the 6xsquare-3-7x and verify the relation ship between the zeroes and the coefficients

Answer 1

$\Large{\underline{\underline{\frak{AnswEr\::}}}}$

We're given a Polynomial 6x² – 7x – 3. & we're asked to find out the zeroes of the Given polynomial and to verify the relationship b/w its zeroes and the coefficients.

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✠ So, Let's find out the Zeroes of a given Polynomial :

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$\dashrightarrow\sf 6x^2 - 7x - 3 = 0\\\\\\\dashrightarrow\sf 6x^2 + 2x - 9x - 3 = 0\\\\\\\dashrightarrow\sf 2x\Big\{3x + 1\Big\} -3\Big\{3x + 1\Big\} = 0\\\\\\\dashrightarrow\sf \Big\{2x - 3\Big\} \;\Big\{3x + 1\Big\} = 0\\\\\\\dashrightarrow{\pmb{\underline{\boxed{\frak{\red{x = \dfrac{3}{2}\;\&-\dfrac{1}{3}}}}}}}\;\bigstar$

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∴ Hence, the zeroes of a given polynomial are ³⁄₂ & –⅓ respectively.

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⌑ On Comparing the given polynomial with standard Quadratic equation (ax² + bx + c = 0), we get —

• a = 6
• b = – 7
• c = – 3

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✇ Now, Verifying the relationship b/w its Zeroes and its Coefficients. As we know that, the Sum & Product of any Quadratic equation are given by :

• (α + β) = -b/a
• (α β) = c/a

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$\quad\quad{\underline{\bf{\dag} \;Sum\;of\;Zeroes\;:}}\\\\\\:\implies\sf \Big\{\alpha + \beta\Big\} = \Bigg\{\dfrac{-b}{a}\Bigg\} \\\\\\:\implies\sf \Bigg\{\dfrac{3}{2} - \dfrac{1}{3}\Bigg\} = \Bigg\{\dfrac{7}{6}\Bigg\}\\\\\\:\implies{\pmb{\boxed{\pink{\frak{\dfrac{7}{6} = \dfrac{7}{6}}}}}}$

$\quad\quad{\underline{\bf{\dag} \; Product\;of\;Zeroes\;:}}\\\\\\:\implies\sf \Big\{\alpha \times \beta\Big\} = \Bigg\{\dfrac{c}{a}\Bigg\} \\\\\\:\implies\sf \Bigg\{\dfrac{3}{2}\times \dfrac{-1 \: }{ \: \: 3 \: }\Bigg\} = \Bigg\{\cancel \dfrac{-3 \: \: }{ \: \: 6 \: \: }\Bigg\}\\\\\\:\implies{\pmb{\boxed{\pink{\frak{\dfrac{-1 \: \: }{ \: \: \: 2 \: \: \: } = \dfrac{-1 \: }{ \: \: 2 \: }}}}}}$

$\quad\quad\quad\;\therefore{\underline{\pmb{\cal{HENCE, \: VERIFIED!}}}}$

Answer 2

Given :-

6x² - 3 - 7x

To Find :-

Zeroes

Solution :-

Rewrite in standard form

6x² - 7x - 3 = 0

6x² - (2x - 9x) - 3 = 0

6x² - 2x + 9x - 3 = 0

2x(3x + 1) - 3(3x + 1) = 0

(3x + 1)(2x - 3) = 0

Either

3x + 1 = 0

3x = 0 - 1

3x = -1

x = -1/3

Or,

2x - 3 = 0

2x = 0 + 3

2x = 3

x = 3/2

Let α = -1/3 and β = 3/2

Let's compare with ax² + bx + c = 0

We observed that

a = 6

b = -7

c = -3

α + β = -b/a

-1/3 + 3/2 = -(-7)/6

-2 + 9/6 = 7/6

7/6 = 7/6

αβ = c/a

-1/3 × 3/2 = -3/6

-1 × 3/3 × 2 = -1/2

-1/2 = -1/2

Hence, Verified as well