Question

Find the zeroes of the polynomial 100x^2 - 81

Answer 1

Hi ,

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

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Since ,

a² - b² = ( a + b )( a - b )

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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 )

I hope this helps you.

: )

Answer 2

Heya !!!



10 × 10 = 100


(10)² = 100



And,


9 × 9 = 81



(9)² = 81



Now,



100X² -81 = 0


(10X)² - (9)² = 0




(10X+9) ( 10X-9) = 0




(10X+9) = 0 OR (10X-9) = 0



10X = -9 OR 10X = 9




X = -9/10 OR X = 9/10.



Hence,



-9/10 and 9/10 are the two zeroes of the given polynomial 100X²-81.



HOPE IT WILL HELP YOU....... :-)