find zero of polynomial 100x^(2) - 81
Answers
Answered by
2
100x² -81 = (10x)² - (9)²
=( 10x+9 ) ( 10x -9 ).......{ a²-b² = (a+b) (a-b) }
Therefore
10x + 9 = 0. or. 10x- 9 = 0
10x = -9. or. 10x = 9
x = -9/10. or. x = 9/ 10
THEREFORE the zeros are -9/10 and 9/10 .
=( 10x+9 ) ( 10x -9 ).......{ a²-b² = (a+b) (a-b) }
Therefore
10x + 9 = 0. or. 10x- 9 = 0
10x = -9. or. 10x = 9
x = -9/10. or. x = 9/ 10
THEREFORE the zeros are -9/10 and 9/10 .
Answered by
1
Answer:
Let p( x ) = 100x² - 81
To find zero of p( x ) , we have to
take p( x ) = 0
100x² - 81 = 0
( 10x )² - 9² = 0
( 10x + 9 )( 10x - 9 ) = 0
Since ,
a² - b² = ( a + b )( a - b )
10x + 9 = 0 or 10x - 9 = 0
10x = -9 or 10x = 9
x = -9/10 or x = 9/10
Therefore ,
-9/10 , 9/10 are zeroes of p( x )
Similar questions