Question

find zeroes of 100x^2-0x-81​

Answer 1

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 )

Answer 2

Answer:

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Step-by-step explanation:

100x^2-81=0

100x^2=81

x^2=81/100

x=√81/100

x=±9/10