Question

Find the zero of the polynomial √2 + 1

Answer 1

$\huge\purple{\mathbb{A᭄ɴsᴡᴇʀ࿐:−}}$

Step-by-step explanation:

Given,

h(s) = 2s2 – (1 + 2√2)s + √2

We put

h(s) = 0 ⇒ 2s2 – (1 + 2√2)s + √2 = 0

⇒ 2s2 – 2√2s – s + √2 = 0 ⇒ 2s(s – √2) -1(s – √2) = 0 ⇒ (2s – 1)(s – √2) = 0

This gives us 2 zeros,

for x = √2 and x = 1/2

Hence, the zeros of the quadratic equation are

√3 and 1.

Now, for verification Sum of zeros = – coefficient of s / coefficient of

s2 √2 + 1/2 = – (-(1 + 2√2)) / 2 (2√2 + 1)/2 = (2√2 +1)/2 Product of roots = constant / coefficient of s2 1/2 x √2 = √2 / 2 √2 / 2 = √2 / 2

Therefore, the relationship between zeros and their coefficients is verified.

Answer 2

$\huge{\bold{\underline{solution:-}}}$

Given:-

$\implies$ Sum of the zeros then,

$\alpha + \beta = \sqrt{2 } + 1$

To Find:-

$\alpha = {?} and \: \beta = {?}$

Now,

we know that ideal equation of polynomial;

$\implies$ ax + b = 0

$\implies$ Put the value of a = √2 and b = 1

Then,

√2 + 1 = 0

On the comparison we got;

$\alpha = \sqrt{2}$

$\beta = 1$

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