Math, asked by nileshmeenaji5178, 3 months ago

please give me full equation solution answer !! urgent
Class 10 th

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Answers

Answered by bryanskiller555
0

Check ur book answer whether my answer is similar to urs or not

Step-by-step explanation:

i).1/4,-1

Given:

Sum of the zeroes=titha+bita=1/4

product of zeroes=titha*bita=-1

The family polynomial is

= k[x^2-(Sum of zeroes)x+(product of zeroes)]

= k[x^2-(1/4)x+(-1)]

= k[x^2-1/4x-1]

= k[4x^2-1x-4 /4]

= k[4x^2-1x-4]

Taking k =1 .°.the required polynomial is

= 1[4x^2-1x-4]

= 4x^2-1x-4

ii)root2,1/3

Given:

sum of the zeroes=titha+bita=root2

product of zeroes=titha*bita=1/3

The family polynomail is

= k[x^2-(Sum of zeroes)x+(product of zeroes)]

= k[x^2-(root2)x+(1/3)]

= k[x^2-(root2)x+1/3]

= k[3x^2-3(root2)x+1 /3]

= k[3x^2-3(root2)x+1]

Taking k =1 .°.the required polynomial is

= 1[3x^2-3(root2)x+1]

= 3x^2-3(root2)x+1

iii)0,root5

Given:

Sum of the zeroes=titha+bita=0

product of zeroes=titha*bita=root5

The family polynomial is

= k[x^2-(Sum of zeroes)x+(product of zeroes)]

= k[x^2-(0)x+(root5)]

= k[x^2-0x+root5]

Taking k =1 .°.the required polynomial is

=1[x^2-0x+root5]

x^2-0x+root5

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