Question

find the zeros of the polynomial 2s2 - [1+2 root 2]s + root 2.

Answer 1

Answer:

Step-by-step explanation:

✒2s²-(1+2√2)s+√2

✒2s²-s-2√2s+√2

✒s(2s-1)-√2(2s-1)

✒(s-√2)(2s-1)

Zeroes are⤵⤵

✏s-√2=0

✏s=√2

________

✏2s-1=0

✏s=½

hope it helps_______SnehaG=======with regàrDs✌✌

Answer 2

Answer:

Zeroes are $\sqrt{2}\:\:and\:\:\frac{1}{2}$

Step-by-step explanation:

Given Quadratic Polynomial , 2s² - ( 1 + 2√2 )s + √2

To find: Zeroes

Consider,

2s² - ( 1 + 2√2 )s + √2 = 0

2s²- s - 2√2s + √2 = 0

s( 2s - 1 ) - √2( 2s - 1 ) = 0

( s - √2 ) ( 2s - 1 ) = 0

s - √2 = 0  and 2s -1 = 0

s = √2  and $s=\frac{1}{2}$

Therefore, Zeroes are $\sqrt{2}\:\:and\:\:\frac{1}{2}$