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Find the zeroes of the polynomial x² + (1/6)x - 2, and verify the relation between the coefficients and the zeroes of the polynomial

Solution:

Given, the polynomial is x² + (1/6)x - 2

We have to find the relation between the coefficients and zeros of the polynomial

The polynomial can be rewritten as \(\frac{1}{6}(6x^{2}+x-12)\)

On factoring,

6x² + x - 12 = 6x² + 9x - 8x - 12

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

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

So, 1/6 (6x² + 6x - 12) = 0

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

3x - 4 = 0

3x = 4

x = 4/3

2x + 3 = 0

2x = -3

x = -3/2

Therefore,the zeros of the polynomial are 4/3 and -3/2.

We know that, if 𝛼 and ꞵ are the zeroes of a polynomial ax² + bx + c, then

Sum of the roots is 𝛼 + ꞵ = -coefficient of x/coefficient of x² = -b/a

Product of the roots is 𝛼ꞵ = constant term/coefficient of x² = c/a

From the given polynomial,

coefficient of x = 1

Coefficient of x² = 6

Constant term = -12

Sum of the roots:

LHS: 𝛼 + ꞵ

=4/3 + (-3/2)

= 8 - 9/6 = -1/6

RHS: -coefficient of x/coefficient of x² = -1/6

LHS = RHS

Product of the roots

LHS: 𝛼ꞵ

= (4/3)(-3/2)

= -12/6 = -2

RHS: constant term/coefficient of x²

= -12/6 = -2.

LHS = RHS

✦ Try This: Find the zeroes of the polynomial x² + (1/7)x - 4, and verify the relation between the coefficients and the zeroes of the polynomial

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 2

NCERT Exemplar Class 10 Maths Exercise 2.3 Sample Problem 1

Find the zeroes of the polynomial x² + (1/6)x - 2, and verify the relation between the coefficients and the zeroes of the polynomial

Summary:

The zeroes of the polynomial x² + (1/6)x - 2 are 4/3 and (-3/2) and the relation between the coefficients and zeros of the polynomial are. Sum of the roots = -b/a = -⅙, Product of the roots = c/a = -2

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