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2s² - (1+ 2√2)s + √2. Find the zeroes of the polynomial, and verify the relation between the coefficients and the zeroes of the polynomial

Solution:

Given, the polynomial is 2s² - (1+ 2√2)s + √2.

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

Let 2s² - (1+ 2√2)s + √2 = 0

On factoring,

2s² - s - 2√2s + √2 = 0

2s² - 2√2s - s + √2 = 0

2s(s - √2 ) - 1(s - √2) = 0

(2s - 1)(s - √2) = 0

Now, 2s - 1 = 0

2s = 1

s = 1/2

Also, s - √2 = 0

s = √2

Therefore,the zeros of the polynomial are 1/2 and √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+2√2)

Coefficient of x² = 2

Constant term = √2

Sum of the roots:

LHS: 𝛼 + ꞵ

= 1/2 + √2

= (1 + 2√2)/2

RHS: -coefficient of x/coefficient of x²

= -[-(1+2√2)/2]

= (1 + 2√2)/2

LHS = RHS

Product of the roots

LHS: 𝛼ꞵ

= (1/2)(√2)

= √2/2

RHS: constant term/coefficient of x²

= √2/2

LHS = RHS

Therefore, the zeros of the polynomial are ½ and √2. The relation between the coefficients is -b/a = (1+2√2)/2 and c/a = √2/2.

✦ Try This: Find the zeroes of the polynomial 3x² + (5√2)x – 3, 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 Problem 7

2s² - (1+ 2√2)s + √2. Find the zeroes of the polynomial, and verify the relation between the coefficients and the zeroes of the polynomial

Summary:

The zeroes of the polynomial 2s² - (1+ 2√2)s + √2 are 1/2 and √2. The relation between the coefficients and zeros of the polynomial are, Sum of the roots = -b/a = (1+2√2)/2, Product of the roots = c/a = √2/2

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