Question

Find the zeroes of the quadratic polynomial 6

2

- 3 -7x and verify the relation

between the zeroes and their coefficients.​

Answer 1

Given :

6x² - 3 - 7x

To find :

• Zero of polynomial
• relationship between zeros and coefficient

Formula used :

For polynomial represented by ax² + bx + c = 0

$\sf \bullet \: \: x \: = \dfrac{ - b \: \pm \: \: \sqrt{ \: {b}^{2} \: - 4ac} }{2a}$

Solution :

Part 1- Zeros of polynomial

Rearrange given polynomial .

6x² - 7x - 3

therefore ,

a = 6

b = -7

c = -3

$\sf \implies \: x \: = \dfrac{ - ( - 7) \: \pm \: \sqrt{( {7}^{2}) \: - \: 4(6)( - 3) } }{2 \times (6)} \\ \\ \\ \sf \implies \: x \: = \frac{7 \: \pm \: \sqrt{ \: 49 \: - ( - 72)} }{12} \\ \\ \sf \implies \: x \: = \dfrac{7 \: \pm \: \sqrt{121} }{12} \\ \\ \sf \implies \: x \: = \dfrac{7 \: \pm \: 11}{12}$

Therefore zeros of given polynomial are

.Either

$\sf \implies \: x \: = \: \dfrac{7 + 11}{12} \\ \\ \sf \implies \: x \: = \dfrac{18}{12} \\ \\ \sf \implies \: x \: = \dfrac{3}{2}$

Or

$\sf \implies \: x \: = \dfrac{7 - 11}{12} \\ \\ \sf \implies \: x \: = \dfrac{ - 4}{12} \\ \\ \sf \implies \: x \: = \dfrac{ - 1}{3}$

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Part 2- Verification of relationship between zeros and coefficient

(1) Sum of zeros = -b/a

LHS ,

$\sf \implies \: Sum \: of \: zeros \: = \: \dfrac{3}{2} + ( \dfrac{ - 1}{3 } ) \\ \\ \sf \implies \: Sum \: of \: zeros \: = \: \dfrac{(3 \times 3) + ( - 1 \times 2)}{3 \times 2} \\ \\ \sf \implies \: Sum \: of \: zeros \: = \: \dfrac{9 - 2}{6} \\ \\ \sf \implies \: Sum \: of \: zeros \: = \: \frac{7}{6}$

RHS , -b/a = -(-7)/6 = (7/6)

LHS = RHS ,

hence sum of zeros = (-b/a) Verified

(2) Products of zeros = c/a

LHS ,

$\sf \implies \: Products \: of \: zeros \: = \dfrac{3}{2} \times \dfrac{ - 1}{3} \\ \\ \sf \implies \: Products \: of \: zeros \: = \dfrac{ - 1}{2}$

RHS , c/a = (-3)/6 = (-1/2)

LHS = RHS ,

HENCE products of zeros = c/a VERIFIED

ANSWER :

1) Zeros of polynomial are , (3/2) & (-1/3)

Answer 2

SOLUTION

TO DETERMINE

• The zeroes of the quadratic polynomial

$\sf{6 {x}^{2} - 3 - 7x}$

• To verify the relation between the zeroes and their coefficients.

EVALUATION

Here the given Quadratic polynomial is

$\sf{6 {x}^{2} - 3 - 7x}$

We are proceeding to find the zeroes of the quadratic polynomial by factorisation method

$\sf{6 {x}^{2} - 3 - 7x}$

$\sf{ = 6 {x}^{2} - 7x - 3}$

$\sf{ = 6 {x}^{2} - (9 - 2)x - 3}$

$\sf{ = 6 {x}^{2} - 9x + 2x - 3}$

$\sf{ = 3x(2x - 3)+1( 2x - 3)}$

$\sf{ = (2x - 3)( 3x + 1)}$

For zeroes of the quadratic polynomial we have

$\sf{6 {x}^{2} - 3 - 7x} = 0$

$\sf{ \implies \: (2x - 3)( 3x + 1) = 0}$

$\displaystyle \sf{ \implies \: \: x = \: \frac{3}{2} \: \: , \: - \frac{1}{3} }$

Hence the zeroes of the quadratic polynomial are

$\displaystyle \sf{ \: \frac{3}{2} \: \: , \: - \frac{1}{3} }$

VERIFICATION

Here the given Quadratic polynomial is

$\sf{6 {x}^{2} - 3 - 7x}$

Comparing with the quadratic polynomial

$\sf{a {x}^{2} + bx + c} \: \: we \: get$

$\sf{a = 6 \: ,\:b = - 7 \: ,\: c = - 3}$

Sum of the zeroes

$\displaystyle \sf{ = \: - \frac{1}{3} + \frac{3}{2}}$

$\displaystyle \sf{ = \: \frac{ - 2 + 9}{6} }$

$\displaystyle \sf{ = \frac{7}{6} }$

$\displaystyle \sf{ Again \: \: - \frac{b}{a} = - \frac{ - 7}{6} = \frac{ 7}{6} }$

$\displaystyle \sf{ \therefore \: \: Sum \: of \: the \: zeroes \: = \: - \frac{b}{a} }$

Product of the Zeroes

$\displaystyle \sf{ = \: - \frac{1}{3} \times \frac{3}{2}}$

$\displaystyle \sf{ = \: - \frac{1}{2} }$

$\displaystyle \sf{ Again \: \: \frac{c}{a} = \frac{ - 3}{6} = \frac{ - 1}{2} }$

$\displaystyle \sf{ Product \: of \: the \: Zeroes \: = \: \frac{c}{a} }$

Hence verified

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