Question

.Find the zeroes of the polynomial x2

+

1

6

x – 2, and verify the relation between the

coefficients and the zeroes of the polynomial.​

Answer 1

Question

Find the zeroes of the polynomial

$\tt{x^{2} + \frac{1}{6}x - 2 }$

and verify the relation between the coefficients and the zeroes of the polynomial.

Answer

Given:

$\large\tt{x^{2} + \frac{1}{6}x - 2 = 0}$

To find : zeroes of polynomial

: verify their relationships

Solution

$\tt{x^{2} + \frac{1}{6}x - 2 = 0 }$

Here, let a=1, b=$\frac{1}{6}$, and c=2

Multiply each term by six, we get

$\tt{ 6x^{2} + x - 12 } = 0$

By splitting the middle terns we get,

6x² +(9-8)x - 12 = 0

=>6x² + 9x + 8x - 12 = 0

=> 3x(2x + 3) - 4(2x +3) = 0

=> (3x-4)(2x+3) = 0

Either,

3x-4 = 0

=> x = $\frac{4}{3}$

Or

2x+3 = 0

=> x = $\frac{-3}{2}$

Hence! the zeroes are $\frac{4}{3}$ and $\frac{-3}{2}$

Verification

Sum of zeroes = $\rm\frac{-b}{a}$

$\mapsto\tt{ \frac{4}{3} + \frac{ - 3} { \: 2} = - \frac{ \frac{1}{6}}{1}}$

$\mapsto \tt{ \frac{8 - 9}{6}} = \frac{ - 1}{ \: 6}$

$\mapsto\tt{ \frac{ - 1}{ \: 6}} = \frac{ - 1}{ \: 6}$

Here, LHS = RHS

Again,

Product of zeroes = $\rm\frac{c}{a}$

$\mapsto\tt{ \frac{4}{3} \times \frac{ - 3}{ \: 2}= \frac{ - 2}{ \: 1}}$

$\mapsto \tt{ \frac{ - 12}{ \: 6} = \frac{ - 2}{ \: 1}}$

$\mapsto\tt{ - 2 = - 2}$

Here too, LHS=RHS.

Hence, verified!

Answer 2

Step-by-step explanation:

Question

Find the zeroes of the polynomial

\tt{x^{2} + \frac{1}{6}x - 2 }x

2

+

6

1

x−2

and verify the relation between the coefficients and the zeroes of the polynomial.

Answer

Given:

\large\tt{x^{2} + \frac{1}{6}x - 2 = 0}x

2

+

6

1

x−2=0

To find : zeroes of polynomial

: verify their relationships

Solution

\tt{x^{2} + \frac{1}{6}x - 2 = 0 }x

2

+

6

1

x−2=0

Here, let a=1, b=\frac{1}{6}

6

1

, and c=2

Multiply each term by six, we get

\tt{ 6x^{2} + x - 12 } = 06x

2

+x−12=0

By splitting the middle terns we get,

6x² +(9-8)x - 12 = 0

=>6x² + 9x + 8x - 12 = 0

=> 3x(2x + 3) - 4(2x +3) = 0

=> (3x-4)(2x+3) = 0

Either,

3x-4 = 0

=> x = \frac{4}{3}

3

4

Or

2x+3 = 0

=> x = \frac{-3}{2}

2

−3

Hence! the zeroes are \frac{4}{3}

3

4

and \frac{-3}{2}

2

−3

Verification

Sum of zeroes = \rm\frac{-b}{a}

a

−b

\mapsto\tt{ \frac{4}{3} + \frac{ - 3} { \: 2} = - \frac{ \frac{1}{6}}{1}}↦

3

4

+

2

−3

=−

1

6

1

\mapsto \tt{ \frac{8 - 9}{6}} = \frac{ - 1}{ \: 6}↦

6

8−9

=

6

−1

\mapsto\tt{ \frac{ - 1}{ \: 6}} = \frac{ - 1}{ \: 6}↦

6

−1

=

6

−1

Here, LHS = RHS

Again,

Product of zeroes = \rm\frac{c}{a}

a

c

\mapsto\tt{ \frac{4}{3} \times \frac{ - 3}{ \: 2}= \frac{ - 2}{ \: 1}}↦

3

4

×

2

−3

=

1

−2

\mapsto \tt{ \frac{ - 12}{ \: 6} = \frac{ - 2}{ \: 1}}↦

6

−12

=

1

−2

\mapsto\tt{ - 2 = - 2}↦−2=−2

Here too, LHS=RHS.

Hence, verified!