plz send me the solution of this question of chapter polynomials
Attachments:
Answers
Answered by
5
Here is your answer (see pic).
Attachments:
cutieboy1:
a big thank s.....
Answered by
1
Hi ,
Let p( x ) = 2x² - ( 1 + 2√2 ) x + √2
To find zeros of p( x ) we have to take
p( x ) = 0
2x² - ( 1 + 2√2 ) x + √2 = 0
2x² - x - 2√2 x + √2 = 0
x ( 2x - 1 ) - √2 ( 2x - 1 ) = 0
( 2x - 1 ) ( x - √2 ) = 0
Therefore ,
2x - 1 = 0 or x - √2 = 0
x = 1/2 or x = √2
1/2 or √2 are Zeroes of p( x ) .
I hope this helps you.
; )
Let p( x ) = 2x² - ( 1 + 2√2 ) x + √2
To find zeros of p( x ) we have to take
p( x ) = 0
2x² - ( 1 + 2√2 ) x + √2 = 0
2x² - x - 2√2 x + √2 = 0
x ( 2x - 1 ) - √2 ( 2x - 1 ) = 0
( 2x - 1 ) ( x - √2 ) = 0
Therefore ,
2x - 1 = 0 or x - √2 = 0
x = 1/2 or x = √2
1/2 or √2 are Zeroes of p( x ) .
I hope this helps you.
; )
Good that you too did by expanding bracket
some fellow were wasting time by using quadratic formula lol
Press red heart
Done
: )
Similar questions